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Inequality Proof Problems [251-300]

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Restructured the IneqMath training data.

P251. (Turkevici’s inequality) Let $a, b, c, d$ be non-negative real numbers. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c, d \in \mathbb{R}^{+}$: \(a^{4}+b^{4}+c^{4}+d^{4}+C a b c d \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}+a^{2} c^{2}+b^{2} d^{2}\) $$

S251. $C = 2$

Without loss of generality we may assume that $a \geq b \geq c \geq d$.\n\nLet us denote\n\(\n\\begin{aligned}\nf(a, b, c, d)= & a^{4}+b^{4}+c^{4}+d^{4}+2 a b c d-a^{2} b^{2}-b^{2} c^{2}-c^{2} d^{2} \\\\\n& -d^{2} a^{2}-a^{2} c^{2}-b^{2} d^{2} \\\\\n= & a^{4}+b^{4}+c^{4}+d^{4}+2 a b c d-a^{2} c^{2}-b^{2} d^{2}-\\left(a^{2}+c^{2}\\right)\\left(b^{2}+d^{2}\\right)\n\\end{aligned}\n\)\n\nWe have\n\(\n\\begin{aligned}\n& f(a, b, c, d)-f(\\sqrt{a c}, b, \\sqrt{a c}, d) \\\\\n&= a^{4}+b^{4}+c^{4}+d^{4}+2 a b c d-a^{2} c^{2}-b^{2} d^{2}-\\left(a^{2}+c^{2}\\right)\\left(b^{2}+d^{2}\\right) \\\\\n&-\\left(a^{2} c^{2}+b^{4}+a^{2} c^{2}+d^{4}+2 a b c d-a^{2} c^{2}-b^{2} d^{2}-2 a c\\left(b^{2}+d^{2}\\right)\\right) \\\\\n&= a^{4}+c^{4}-2 a^{2} c^{2}-\\left(b^{2}+d^{2}\\right)\\left(a^{2}+c^{2}+2 a c\\right) \\\\\n&=\\left(a^{2}-c^{2}\\right)^{2}-\\left(b^{2}+d^{2}\\right)(a-c)^{2}=(a-c)^{2}\\left((a+c)^{2}-\\left(b^{2}+d^{2}\\right)\\right) \\geq 0\n\\end{aligned}\n\)\n\nThus\n\(\nf(a, b, c, d) \\geq f(\\sqrt{a c}, b, \\sqrt{a c}, d)\n\)\n\nBy the SMV theorem we only need to prove that $f(a, b, c, d) \geq 0$, in the case when $a=b=c=t \geq d$.\n\nWe have\n$f(t, t, t, d) \geq 0 \quad \Leftrightarrow \quad 3 t^{4}+d^{4}+2 t^{3} d \geq 3 t^{4}+3 t^{2} d^{2} \quad \Leftrightarrow \quad d^{4}+2 t^{3} d \geq 3 t^{2} d^{2}$, which immediately follows from $A M \geq G M$.\n\nEquality occurs when $a = b = c = d$ or when $a = b = c$ and $d = 0$ (up to permutation). In both cases, the inequality becomes an equality, so $C = 2$ is the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.

S251.1.

S251.2. \(\frac{3}{2}\,(a^4+b^4+c^4+d^4)\ \ge\ a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2\) can be understood easily (e.g., by the rearrangement inequality). Also, \(6abcd\ \le\ a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2\) can be understood easily (e.g., by the rearrangement inequality). Therefore, the given problem becomes: determine the critical value \(p\) for which \(\frac{6-p}{4}\,(a^4+b^4+c^4+d^4) + p\,abcd \ \ge\ a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2\) is a tight inequality for some \(p\). (This can be seen easily from the fact that when \(p\) is small, the left-hand side is larger, and when \(p\) is large, the right-hand side is larger.) Thus, we can reduce it to the problem of determining that critical value \(p\). If \(a=b=c=d\), we cannot obtain any condition on \(p\). If one of \(a,b,c,d\) is \(0\) and the other three are all equal, we can conclude that \(p\le 2\). Hence, by the existence of a tight inequality, the critical value \(p\) is determined either at a boundary case or at an interior extremal point, so we get \(p=2.\)

P252. Let $P, L, R$ denote the area, perimeter, and circumradius of $\triangle ABC$, respectively. Find the smallest constant $C$ such that the following inequality holds for all triangles $\triangle ABC$: \(\frac{L P}{R^3} \leq C.\)

S252. $C = \frac{27}{4}$

We have\n\(\n\\frac{L P}{R^{3}}=\\frac{(a+b+c) a b c}{R^{3} 4 R}=\\frac{2 R(\\sin \\alpha+\\sin \\beta+\\sin \\gamma) 8 R^{3} \\sin \\alpha \\sin \\beta \\sin \\gamma}{4 R^{4}}\n\)\ni.e.\n\(\n\\begin{equation*}\n\\frac{L P}{R^{3}}=4(\\sin \\alpha+\\sin \\beta+\\sin \\gamma) \\sin \\alpha \\sin \\beta \\sin \\gamma \\tag{1}\n\\end{equation*}\n\)\n\nBy $A M \geq G M$ we have\n\(\n\\sin \\alpha \\sin \\beta \\sin \\gamma \\leq\\left(\\frac{\\sin \\alpha+\\sin \\beta+\\sin \\gamma}{3}\\right)^{3}\n\)\n\nSo by (1) we get\n\(\n\\begin{equation*}\n\\frac{L P}{R^{3}} \\leq \\frac{4(\\sin \\alpha+\\sin \\beta+\\sin \\gamma)^{4}}{27} \\tag{2}\n\\end{equation*}\n\)\n\nThe function $f(x)=-\sin x$ is convex on $[0, \pi]$, so by Jensen’s inequality we have\n\(\n\\frac{\\sin \\alpha+\\sin \\beta+\\sin \\gamma}{3} \\leq \\sin \\left(\\frac{\\alpha+\\beta+\\gamma}{3}\\right)=\\frac{\\sqrt{3}}{2}\n\)\n\nFinally from (2) we obtain\n\(\n\\frac{L P}{R^{3}} \\leq \\frac{4}{27}\\left(\\frac{3 \\sqrt{3}}{2}\\right)^{4}=\\frac{27}{4}\n\)\n\nEquality holds when $a = b = c$, i.e., when the triangle is equilateral. In this case, the maximum value of $\frac{L P}{R^3}$ is achieved, so $C = \frac{27}{4}$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{27}{4}$.

P253. Let $a, b, c, d$ be positive real numbers such that $a b c d = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c, d$ satisfying the given constraint: \(\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+d)}+\frac{1}{d(1+a)} \geq C\)

S253. $C = 2$

With the substitutions $a=\frac{x}{y}, b=\frac{t}{x}, c=\frac{z}{t}, d=\frac{y}{z}$, the given inequality becomes\n\(\n\\frac{x}{z+t}+\\frac{y}{x+t}+\\frac{z}{x+y}+\\frac{t}{z+y} \\geq 2\n\)\n\nBy the Cauchy-Schwarz inequality we have\n\(\n\\begin{aligned}\n\\frac{x}{z+t}+\\frac{y}{x+t}+\\frac{z}{x+y}+\\frac{t}{z+y} & =\\frac{x^{2}}{x z+x t}+\\frac{y^{2}}{y x+y t}+\\frac{z^{2}}{z x+z y}+\\frac{t^{2}}{t z+t y} \\\\\n& \\geq \\frac{(x+y+z+t)^{2}}{2 x z+2 y t+x t+y x+z y+t z}\n\\end{aligned}\n\)\n\nHence it suffices to prove that\n\(\n\\frac{(x+y+z+t)^{2}}{2 x z+2 y t+x t+y x+z y+t z} \\geq 2\n\)\nwhich is equivalent to\n\(\n(x-z)^{2}+(y-t)^{2} \\geq 0\n\)\n\nEquality occurs when $x = z$ and $y = t$, which corresponds to $a = c = 1/b = 1/d$. In this case, the original sum achieves its minimum value. Thus, $C = 2$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.

P254. Let $x, y, z$ be real numbers. Find the largest constant $C$ such that the following inequality holds for all $x, y, z \in \mathbb{R}$: \(x^{4}+y^{4}+z^{4} \geq 4 x y z + C\)

S254. $C = -1$

We have\n\(\n\\begin{aligned}\nx^{4} & +y^{4}+z^{4}-4 x y z+1 \\\\\n& =\\left(x^{4}-2 x^{2}+1\\right)+\\left(y^{4}-2 y^{2} z^{2}+z^{4}\\right)+\\left(2 y^{2} z^{2}-4 x y z+2 x^{2}\\right) \\\\\n& =\\left(x^{2}-1\\right)^{2}+\\left(y^{2}-z^{2}\\right)^{2}+2(y z-x)^{2} \\geq 0\n\\end{aligned}\n\)\nso it follows that\n\(\nx^{4}+y^{4}+z^{4} \\geq 4 x y z-1\n\)\n\nEquality holds when $x = 1$, $y = z = 0$, or any permutation thereof, since then all squared terms vanish. This gives the minimum value of $C$, so $C = -1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = -1$.

P255. Let $a, b, c$ be real numbers different from 1, such that $a+b+c=1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(\frac{1+a^{2}}{1-a^{2}}+\frac{1+b^{2}}{1-b^{2}}+\frac{1+c^{2}}{1-c^{2}} \geq C\)

S255. $C = \frac{15}{4}$

Since $a, b, c>0, a \neq 1, b \neq 1, c \neq 1$ and $a+b+c=1$ it follows that $0<a, b, c<1$.\n\nThe given inequality is symmetric, so without loss of generality we may assume that $a \leq b \leq c$.\n\nThen we have\n\(\n1+a^{2} \\leq 1+b^{2} \\leq 1+c^{2} \\quad \\text { and } \\quad 1-c^{2} \\leq 1-b^{2} \\leq 1-a^{2}\n\)\n\nHence\n\(\n\\frac{1}{1-a^{2}} \\leq \\frac{1}{1-b^{2}} \\leq \\frac{1}{1-c^{2}}\n\)\n\nNow by Chebishev’s inequality we have\n\(\n\\begin{aligned}\nA & =\\frac{1+a^{2}}{1-a^{2}}+\\frac{1+b^{2}}{1-b^{2}}+\\frac{1+c^{2}}{1-c^{2}} \\\\\n& \\geq \\frac{1}{3}\\left(1+a^{2}+1+b^{2}+1+c^{2}\\right)\\left(\\frac{1}{1-a^{2}}+\\frac{1}{1-b^{2}}+\\frac{1}{1-c^{2}}\\right)\n\\end{aligned}\n\)\ni.e.\n\(\n\\begin{equation*}\nA \\geq \\frac{\\left(a^{2}+b^{2}+c^{2}+3\\right)}{3}\\left(\\frac{1}{1-a^{2}}+\\frac{1}{1-b^{2}}+\\frac{1}{1-c^{2}}\\right) \\tag{1}\n\\end{equation*}\n\)\n\nAlso we have the well-known inequality\n\(\na^{2}+b^{2}+c^{2} \\geq \\frac{(a+b+c)^{2}}{3}=\\frac{1}{3}\n\)\n\nTherefore by (1) we obtain\n\(\n\\begin{equation*}\nA \\geq \\frac{(1 / 3+3)}{3}\\left(\\frac{1}{1-a^{2}}+\\frac{1}{1-b^{2}}+\\frac{1}{1-c^{2}}\\right)=\\frac{10}{9}\\left(\\frac{1}{1-a^{2}}+\\frac{1}{1-b^{2}}+\\frac{1}{1-c^{2}}\\right) \\tag{2}\n\\end{equation*}\n\)\n\nSince $1-a^{2}, 1-b^{2}, 1-c^{2}>0$, by using $A M \geq H M$ we deduce\n\(\n\\begin{equation*}\n\\frac{1}{1-a^{2}}+\\frac{1}{1-b^{2}}+\\frac{1}{1-c^{2}} \\geq \\frac{9}{3-\\left(a^{2}+b^{2}+c^{2}\\right)} \\geq \\frac{9}{3-1 / 3}=\\frac{27}{8} \\tag{3}\n\\end{equation*}\n\)\n\nFinally from (2) and (3) we get\n\(\nA \\geq \\frac{10}{9}\\left(\\frac{1}{1-a^{2}}+\\frac{1}{1-b^{2}}+\\frac{1}{1-c^{2}}\\right) \\geq \\frac{10}{9} \\cdot \\frac{27}{8}=\\frac{15}{4}\n\)\nwith equality iff $a=b=c=1 / 3$.\n\nEquality is achieved when $a = b = c = \frac{1}{3}$, which satisfies $a + b + c = 1$ and $0 < a, b, c < 1$. In this case, the minimum value of the given expression is $\frac{15}{4}$, so $C = \frac{15}{4}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{15}{4}$.

P256. Let $a, b, c \in \mathbb{R}^{+}$. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$:

S256. $C = \frac{1}{81}$

We have\n\(\n\\begin{aligned}\n&(1+a)(a+b)(b+c)(c+16) \\\\\n&=\\left(1+\\frac{a}{2}+\\frac{a}{2}\\right)\\left(a+\\frac{b}{2}+\\frac{b}{2}\\right)\\left(b+\\frac{c}{2}+\\frac{c}{2}\\right)(c+8+8) \\\\\n& \\geq 3 \\sqrt[3]{\\frac{a^{2}}{4}} \\cdot 3 \\sqrt[3]{\\frac{a b^{2}}{4}} \\cdot 3 \\sqrt[3]{\\frac{b c^{2}}{4}} \\cdot 3 \\sqrt[3]{\\frac{64 c}{4}} \\geq 81 a b c\n\\end{aligned}\n\)\n\nThus\n\(\n\\frac{a b c}{(1+a)(a+b)(b+c)(c+16)} \\leq \\frac{1}{81}\n\)\n\nEquality holds when all the terms inside the cube roots are equal, which occurs when $a = b = c = 1$. In this case, the expression attains its maximum value, so $C = \frac{1}{81}$ is the minimal constant such that the inequality always holds.\n\nTherefore, the answer is $C = \frac{1}{81}$.

P257. Let $a, b, c \in (1,2)$ be real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in (1,2)$: \(\frac{b \sqrt{a}}{4 b \sqrt{c}-c \sqrt{a}}+\frac{c \sqrt{b}}{4 c \sqrt{a}-a \sqrt{b}}+\frac{a \sqrt{c}}{4 a \sqrt{b}-b \sqrt{c}} \geq C\)

S257. $C = 1$

Since $a, b, c \in(1,2)$ we have\n\(\n4 b \\sqrt{c}-c \\sqrt{a}>4 \\sqrt{c}-2 \\sqrt{c}=2 \\sqrt{c}>0\n\)\n\nAnalogously we get $4 c \sqrt{a}-a \sqrt{b}>0$ and $4 a \sqrt{b}-b \sqrt{c}>0$.\nWe’ll prove that\n\(\n\\begin{equation*}\n\\frac{b \\sqrt{a}}{4 b \\sqrt{c}-c \\sqrt{a}} \\geq \\frac{a}{a+b+c} \\tag{1}\n\\end{equation*}\n\)\n\nSince $4 b \sqrt{c}-c \sqrt{a}>0$ inequality (1) is\n\(\n\\begin{array}{r}\nb(a+b+c) \\geq \\sqrt{a}(4 b \\sqrt{c}-c \\sqrt{a}) \\\\\n\\Leftrightarrow \\quad(a+b)(b+c) \\geq 4 b \\sqrt{a c}\n\\end{array}\n\)\nwhich is clearly true $(A M \geq G M)$.\nSimilarly we deduce that\n\(\n\\begin{equation*}\n\\frac{c \\sqrt{b}}{4 c \\sqrt{a}-a \\sqrt{b}} \\geq \\frac{b}{a+b+c} \\tag{2}\n\\end{equation*}\n\)\nand\n\(\n\\begin{equation*}\n\\frac{a \\sqrt{c}}{4 a \\sqrt{b}-b \\sqrt{c}} \\geq \\frac{c}{a+b+c} \\tag{3}\n\\end{equation*}\n\)\n\nAdding (1), (2) and (3) we get the required result.\n\nEquality holds when $a = b = c$, and since $a, b, c \in (1,2)$, this occurs for any $a = b = c$ in $(1,2)$. In this case, each term becomes $\frac{a}{a+b+c} = \frac{1}{3}$, so the sum is $1$. This gives the minimum value of the expression, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P258. Let $a, b, c$ be positive real numbers such that $a b c = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c$: \(\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geq C\)

S258. $C = \frac{3}{2}$

Without loss of generality we may assume that $a \geq b \geq c$.\nLet $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$. Then clearly $x y z=1$.\n\nWe have\n\(\n\\begin{aligned}\n\\frac{1}{a^{3}(b+c)}+\\frac{1}{b^{3}(c+a)}+\\frac{1}{c^{3}(a+b)} & =\\frac{x^{3}}{1 / y+1 / z}+\\frac{y^{3}}{1 / z+1 / x}+\\frac{z^{3}}{1 / x+1 / y} \\\\\n& =\\frac{x^{2}}{y+z}+\\frac{y^{2}}{z+x}+\\frac{z^{2}}{x+y}\n\\end{aligned}\n\)\n\nSince $c \leq b \leq a$ we have $x \leq y \leq z$.\nSo clearly $x+y \leq z+x \leq y+z$ and $\frac{x}{y+z} \leq \frac{y}{z+x} \leq \frac{z}{x+y}$.\nNow by the rearrangement inequality we get the following inequalities\n\(\n\\begin{aligned}\n& \\frac{x^{2}}{y+z}+\\frac{y^{2}}{z+x}+\\frac{z^{2}}{x+y} \\geq \\frac{x y}{y+z}+\\frac{y z}{z+x}+\\frac{z x}{x+y} \\\\\n& \\frac{x^{2}}{y+z}+\\frac{y^{2}}{z+x}+\\frac{z^{2}}{x+y} \\geq \\frac{x z}{y+z}+\\frac{y x}{z+x}+\\frac{z y}{x+y}\n\\end{aligned}\n\)\n\nSo we obtain\n\(\n\\begin{aligned}\n& 2\\left(\\frac{1}{a^{3}(b+c)}+\\frac{1}{b^{3}(c+a)}+\\frac{1}{c^{3}(a+b)}\\right) \\\\\n& \\quad=2\\left(\\frac{x^{2}}{y+z}+\\frac{y^{2}}{z+x}+\\frac{z^{2}}{x+y}\\right) \\\\\n& \\quad \\geq \\frac{x y}{y+z}+\\frac{y z}{z+x}+\\frac{z x}{x+y}+\\frac{x z}{y+z}+\\frac{y x}{z+x}+\\frac{z y}{x+y} \\\\\n& \\quad=x+y+z \\geq 3 \\sqrt[3]{x y z}=3\n\\end{aligned}\n\)\nas required.\n\nEquality holds when $a = b = c = 1$, which implies $x = y = z = 1$. In this case, the sum becomes $3 \times \frac{1}{1^3(1+1)} = 3 \times \frac{1}{2} = \frac{3}{2}$. Thus, the minimum value of the expression is $\frac{3}{2}$, so $C = \frac{3}{2}$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{3}{2}$.

P259. Find the smallest constant $C$ such that for all real numbers $x$, the following inequality holds: \(2 x^{4} + C \geq 2 x^{3} + x^{2}\)

S259. $C = 1$

We have\n\(\n\\begin{aligned}\n2 x^{4}+1-2 x^{3}-x^{2} & =1-x^{2}-2 x^{3}(1-x)=(1-x)(1+x)-2 x^{3}(1-x) \\\\\n& =(1-x)\\left(x+1-2 x^{3}\\right)=(1-x)\\left(x\\left(1-x^{2}\\right)+1-x^{3}\\right) \\\\\n& =(1-x)\\left(x(1-x)(1+x)+(1-x)\\left(1+x+x^{2}\\right)\\right) \\\\\n& =(1-x)\\left((1-x)\\left(x(1+x)+1+x+x^{2}\\right)\\right) \\\\\n& =(1-x)^{2}\\left((x+1)^{2}+x^{2}\\right) \\geq 0 .\n\\end{aligned}\n\)\n\nEquality occurs if and only if $x = 1$, since then $(1-x)^2 = 0$. This gives the minimum value of $C$, so $C = 1$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P260. Let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$: \(\frac{a^{3}+2}{b+2}+\frac{b^{3}+2}{c+2}+\frac{c^{3}+2}{a+2} \geq C.\)

S260. $C = 3$

By $A M \geq G M$ we have\n\(\n\\frac{a^{3}+2}{b+2}=\\frac{a^{3}+1+1}{b+2} \\geq \\frac{3 \\sqrt[3]{a^{3} \\cdot 1 \\cdot 1}}{b+2}=\\frac{3 a}{b+2}\n\)\n\nSimilarly we get\n\(\n\\frac{b^{3}+2}{c+2} \\geq \\frac{3 b}{c+2} \\quad \\text { and } \\quad \\frac{c^{3}+2}{a+2} \\geq \\frac{3 c}{a+2}\n\)\n\nTherefore\n\(\n\\begin{equation*}\n\\frac{a^{3}+2}{b+2}+\\frac{b^{3}+2}{c+2}+\\frac{c^{3}+2}{a+2} \\geq 3\\left(\\frac{a}{b+2}+\\frac{b}{c+2}+\\frac{c}{a+2}\\right) \\tag{1}\n\\end{equation*}\n\)\n\nApplying the Cauchy-Schwarz inequality we obtain\n\(\n\\begin{align*}\n\\frac{a}{b+2}+\\frac{b}{c+2}+\\frac{c}{a+2} & =\\frac{a^{2}}{a(b+2)}+\\frac{b^{2}}{b(c+2)}+\\frac{c^{2}}{c(a+2)} \\\\\n& \\geq \\frac{(a+b+c)^{2}}{a(b+2)+b(c+2)+c(a+2)} \\\\\n& =\\frac{(a+b+c)^{2}}{a b+b c+c a+2(a+b+c)} \\tag{2}\n\\end{align*}\n\)\n\nSince $(a+b+c)^{2} \geq 3(a b+b c+c a)$ we deduce that\n\(\n\\begin{equation*}\n\\frac{1}{a b+b c+c a} \\geq \\frac{3}{(a+b+c)^{2}} \\tag{3}\n\\end{equation*}\n\)\n\nFrom (2) and (3) we get\n\(\n\\begin{align*}\n\\frac{a}{b+2}+\\frac{b}{c+2}+\\frac{c}{a+2} & \\geq \\frac{(a+b+c)^{2}}{a b+b c+c a+2(a+b+c)} \\\\\n& \\geq \\frac{(a+b+c)^{2}}{(a+b+c)^{2} / 3+2(a+b+c)} \\\\\n& =\\frac{3(a+b+c)^{2}}{(a+b+c)^{2}+6(a+b+c)}=\\frac{3(a+b+c)}{(a+b+c)+6} \\tag{4}\n\\end{align*}\n\)\n\nFinally by (1), (4) and since $a+b+c=3$ we obtain\n\(\nA \\geq 3\\left(\\frac{a}{b+2}+\\frac{b}{c+2}+\\frac{c}{a+2}\\right) \\geq \\frac{9(a+b+c)}{(a+b+c)+6}=\\frac{27}{9}=3\n\)\nas required. Equality occurs iff $a=b=c=1$.\n\nEquality holds when $a = b = c = 1$, since then $a + b + c = 3$ and each term becomes $\frac{1^3 + 2}{1 + 2} = 1$, so the sum is $3$. Thus, the minimum value of the expression is $3$, so the largest constant $C$ is $3$.\n\nTherefore, the answer is $C = 3$.

P261. Let $a, b, c$ be the lengths of the sides of a triangle, and let $l_{\alpha}, l_{\beta}, l_{\gamma}$ be the lengths of the bisectors of the respective angles. Let $s$ be the semi-perimeter and $r$ denote the inradius of the triangle. Determine the minimal constant $C$ such that the following inequality holds for all triangles: \(\frac{l_{\alpha}}{a}+\frac{l_{\beta}}{b}+\frac{l_{\gamma}}{c} \leq C\frac{s}{r}.\)

S261. $C = \frac{1}{2}$

The following identities hold:\n\(\nl_{\\alpha}=\\frac{2 \\sqrt{b c}}{b+c} \\sqrt{s(s-a)}, \\quad l_{\\beta}=\\frac{2 \\sqrt{c a}}{c+a} \\sqrt{s(s-b)} \\quad \\text { and } \\quad l_{\\gamma}=\\frac{2 \\sqrt{a b}}{a+b} \\sqrt{s(s-c)}\n\)\n\nFrom the obvious inequality $\frac{2 \sqrt{x y}}{x+y} \leq 1$ and the previous identities we obtain that\n\(\n\\begin{equation*}\nl_{\\alpha} \\leq \\sqrt{s(s-a)}, \\quad l_{\\beta} \\leq \\sqrt{s(s-b)} \\quad \\text { and } \\quad l_{\\gamma} \\leq \\sqrt{s(s-c)} \\tag{1}\n\\end{equation*}\n\)\n\nAlso\n\(\n\\begin{equation*}\nh_{a} \\leq l_{\\alpha}, \\quad h_{b} \\leq l_{\\beta} \\quad \\text { and } \\quad h_{c} \\leq l_{\\gamma} \\tag{2}\n\\end{equation*}\n\)\n\nSo we have\n\(\n\\begin{aligned}\n\\frac{l_{\\alpha}}{a}+\\frac{l_{\\beta}}{b}+\\frac{l_{\\gamma}}{c} & =\\frac{l_{\\alpha} h_{a}}{2 P}+\\frac{l_{\\beta} h_{b}}{2 P}+\\frac{l_{\\gamma} h_{c}}{2 P} \\stackrel{(2)}{\\leq} \\frac{l_{\\alpha}^{2}+l_{\\beta}^{2}+l_{\\gamma}^{2}}{2 P} \\\\\n& \\stackrel{(1)}{\\leq} \\frac{s(s-a)+s(s-b)+s(s-c)}{2 P} \\\\\n& =\\frac{3 s^{2}-s(a+b+c)}{2 r s}=\\frac{3 s^{2}-2 s^{2}}{2 r s}=\\frac{s^{2}}{2 r s}=\\frac{s}{2 r}\n\\end{aligned}\n\)\n\nEquality occurs if and only if the triangle is equilateral, since all the inequalities above become equalities only in this case. This gives the maximum value of $\frac{l_{\alpha}}{a}+\frac{l_{\beta}}{b}+\frac{l_{\gamma}}{c}$ relative to $\frac{s}{r}$, so the minimal constant $C$ is $\frac{1}{2}$.\n\nTherefore, the answer is $C = \frac{1}{2}$.

P262. Let $a, b, c$ be the side lengths of a given triangle. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the triangle inequality: \(a^{2}+b^{2}+c^{2} < C(a b+b c+c a)\)

S262. $C = 2$

Let $a=x+y, b=y+z, c=z+x, x, y, z>0$.\nThen we have\n\(\n\\begin{aligned}\n& (x+y)^{2}+(y+z)^{2}+(z+x)^{2} \\\\\n& \\quad<2((x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y))\n\\end{aligned}\n\)\nor\n\(\nxy+yz+zx>0\n\)\nwhich is clearly true.\n\nEquality is approached as $x, y, z \to 0$ (i.e., when the triangle becomes degenerate), but never actually achieved for positive $x, y, z$. Thus, $C = 2$ is the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.

P263. Let $a, b, c$ be positive real numbers such that $a + b + c = 6$. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(\sqrt[3]{a b + b c} + \sqrt[3]{b c + c a} + \sqrt[3]{c a + a b} \leq C\)

S263. $C = 6$

By the power mean inequality we have\n\(\n\\frac{\\sqrt[3]{a b+b c}+\\sqrt[3]{b c+c a}+\\sqrt[3]{c a+a b}}{3} \\leq \\sqrt[3]{\\frac{(a b+b c)+(b c+c a)+(c a+a b)}{3}}\n\)\ni.e.\n\(\n\\begin{equation*}\n\\sqrt[3]{a b+b c}+\\sqrt[3]{b c+c a}+\\sqrt[3]{c a+a b} \\leq \\sqrt[3]{18(a b+b c+c a)} \\tag{1}\n\\end{equation*}\n\)\n\nSince $a b+b c+c a \leq \frac{(a+b+c)^{2}}{3}=12$ by (1) we obtain\n\(\n\\sqrt[3]{a b+b c}+\\sqrt[3]{b c+c a}+\\sqrt[3]{c a+a b} \\leq \\sqrt[3]{18 \\cdot 12}=6\n\)\n\nEquality occurs if and only if $a = b = c = 2$, which is when $a, b, c$ are all equal and $a + b + c = 6$. In this case, the maximum value of the expression is achieved, so $C = 6$ is the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 6$.

P264. Let $n \geq 2, n \in \mathbb{N}$ and $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers such that \(\frac{1}{x_{1}+1998}+\frac{1}{x_{2}+1998}+\cdots+\frac{1}{x_{n}+1998}=\frac{1}{1998}\)

Find the largest constant $C$ such that the following inequality holds for all $x_{1}, x_{2}, \ldots, x_{n}$ satisfying the given constraint: \(\sqrt[n]{x_{1} x_{2} \cdots x_{n}} \geq C\)

S264. $C = 1998(n-1)$

After setting $\frac{1998}{x_{i}+1998}=a_{i}$, for $i=1,2, \ldots, n$, the identity\n\(\n\\frac{1}{x_{1}+1998}+\\frac{1}{x_{2}+1998}+\\cdots+\\frac{1}{x_{n}+1998}=\\frac{1}{1998}\n\)\nbecomes\n\(\na_{1}+a_{2}+\\cdots+a_{n}=1\n\)\n\nWe need to show that\n\(\n\\begin{equation*}\n\\left(\\frac{1}{a_{1}}-1\\right)\\left(\\frac{1}{a_{2}}-1\\right) \\cdots\\left(\\frac{1}{a_{n}}-1\\right) \\geq(n-1)^{n} \\tag{1}\n\\end{equation*}\n\)\n\nWe have\n\(\n\\begin{aligned}\n\\frac{1}{a_{i}}-1 & =\\frac{1-a_{i}}{a_{i}}=\\frac{a_{1}+\\cdots+a_{i-1}+a_{i+1}+\\cdots+a_{n}}{a_{i}} \\\\\n& \\geq(n-1) \\sqrt[n-1]{\\frac{a_{1} \\cdots a_{i-1} a_{i+1} \\cdots a_{n}}{a_{i}^{n-1}}}\n\\end{aligned}\n\)\n\nMultiplying these inequalities for $i=1,2, \ldots, n$ we obtain (1), as desired.\n\nEquality holds when all $a_i$ are equal, i.e., $a_1 = a_2 = \cdots = a_n = \frac{1}{n}$. This corresponds to $x_i = 1998(n-1)$ for all $i$. Thus, the minimum value of $\sqrt[n]{x_1 x_2 \cdots x_n}$ is achieved at this point, and $C = 1998(n-1)$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1998(n-1)$.

P265. Let $a, b, c, d$ be positive real numbers such that $abcd = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c, d$ satisfying the given constraint: \(\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq C\)

S265. $C = 1$

If follows by summing the inequalities\n\n\(\n\\begin{aligned}\n& \\frac{1}{(1+a)^{2}}+\\frac{1}{(1+b)^{2}} \\geq \\frac{1}{1+a b} \\\\\n& \\frac{1}{(1+c)^{2}}+\\frac{1}{(1+d)^{2}} \\geq \\frac{1}{1+c d}\n\\end{aligned}\n\)\n\nThe first from these inequalities follows from\n\n\(\n\\begin{aligned}\n\\frac{1}{(1+a)^{2}}+\\frac{1}{(1+b)^{2}}-\\frac{1}{1+a b} & =\\frac{a b\\left(a^{2}+b^{2}\\right)-a^{2} b^{2}-2 a b+1}{(1+a)^{2}(1+b)^{2}(1+c)^{2}}= \\\\\n& =\\frac{a b(a-b)^{2}+(a b-1)^{2}}{(1+a)^{2}(1+b)^{2}(1+a b)} \\geq 0\n\\end{aligned}\n\)\n\nEquality holds if $a=b=c=d=1$.\n\nEquality in the above inequalities is achieved when $a = b = c = d = 1$, which also satisfies the constraint $abcd = 1$. In this case, each term is $\frac{1}{4}$, so the sum is $1$. This gives the minimum value of the sum, so the maximal constant $C$ is $1$.\n\nTherefore, the answer is $C = 1$.

P266. Let $x, y, z$ be real numbers different from 1, such that $x y z = 1$. Find the largest constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint: \(\left(\frac{3-x}{1-x}\right)^{2}+\left(\frac{3-y}{1-y}\right)^{2}+\left(\frac{3-z}{1-z}\right)^{2} > C\)

S266. $C = 7$

Denote $A=\left(\frac{3-x}{1-x}\right)^{2}+\left(\frac{3-y}{1-y}\right)^{2}+\left(\frac{3-z}{1-z}\right)^{2}-7$.\nWe have\n\(\nA=\\left(1+\\frac{2}{1-x}\\right)^{2}+\\left(1+\\frac{2}{1-y}\\right)^{2}+\\left(1+\\frac{2}{1-z}\\right)^{2}-7\n\)\n\nLet $\frac{1}{1-x}=a, \frac{1}{1-y}=b, \frac{1}{1-z}=c$.\nThen $A=(1+2 a)^{2}+(1+2 b)^{2}+(1+2 c)^{2}-7$, i.e.\n\(\n\\begin{equation*}\nA=4 a^{2}+4 b^{2}+4 c^{2}+4 a+4 b+4 c-4 \\tag{1}\n\\end{equation*}\n\)\n\nFurthermore, the condition $x y z=1$ is equivalent to $a b c=(a-1)(b-1)(c-1)$, i.e.\n\(\n\\begin{equation*}\na+b+c-1=a b+b c+c a \\tag{2}\n\\end{equation*}\n\)\n\nUsing (1) and (2) we get\n\(\nA=4 a^{2}+4 b^{2}+4 c^{2}+4(a b+b c+c a)=2\\left((a+b)^{2}+(b+c)^{2}+(c+a)^{2}\\right)\n\)\ni.e. $A \geq 0$.\n\nEquality occurs if and only if $a=b=c=0$, which is impossible under the given constraints. Therefore, the minimum value of the original expression is strictly greater than $7$, and $C = 7$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 7$.

P267. Let $a, b \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all $a, b \in \mathbb{R}^{+}$: \(C(a^3 + b^3) \geq (a + b)^3\)

S267. $C = 4$

The function $f(x)=x^{3}$ is convex on $(0,+\infty)$, thus from Jensen’s inequality it follows that\n\(\n\\left(\\frac{a+b}{2}\\right)^{3} \\leq \\frac{a^{3}+b^{3}}{2} \\Leftrightarrow 4\\left(a^{3}+b^{3}\\right) \\geq(a+b)^{3}\n\)\n\nEquality holds when $a = b$, since then $a^3 + b^3 = 2a^3$ and $(a + b)^3 = (2a)^3 = 8a^3$, so $4(a^3 + b^3) = 8a^3 = (a + b)^3$. This gives the minimum value of $C$, so $C = 4$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 4$.

P268. Let $x, y, z > 0$ be real numbers. Determine the maximal constant $C$ such that the following inequality holds for all $x, y, z$: \((x y + y z + z x) \left( \frac{1}{(x + y)^2} + \frac{1}{(y + z)^2} + \frac{1}{(z + x)^2} \right) \geq C.\)

S268. $C = \frac{9}{4}$

The given inequality is equivalent to\n\(\n\\begin{align*}\n& 4(x y+y z+z x)\\left((z+x)^{2}(y+z)^{2}+(x+y)^{2}(z+x)^{2}+(x+y)^{2}(y+z)^{2}\\right) \\\\\n& \\quad \\geq 9(x+y)^{2}(y+z)^{2}(z+x)^{2} \\tag{1}\n\\end{align*}\n\)\n\nLet us denote $p=x+y+z, q=x y+y z+z x, r=x y z$.\nBy $I_{5}$ and $I_{7}$ we have\n\(\n(x+y)^{2}(y+z)^{2}(z+x)^{2}=(p q-r)^{2}\n\)\nand\n\(\n(x+y)^{2}(y+z)^{2}+(y+z)^{2}(z+x)^{2}+(z+x)^{2}(x+y)^{2}=\\left(p^{2}+q\\right)^{2}-4 p(p q-r)\n\)\n\nSo we can rewrite inequality (1) as follows\n\(\n\\begin{aligned}\n& 4 q\\left(\\left(p^{2}+q\\right)^{2}-4 p(p q-r)\\right) \\geq 9(p q-r)^{2} \\\\\n& \\quad \\Leftrightarrow \\quad 4 p^{4} q-17 p^{2} q^{2}+4 q^{3}+34 p q r-9 r^{2} \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad 3 p q\\left(p^{3}-4 p q+9 r\\right)+q\\left(p^{4}-5 p^{2} q+4 q^{2}+6 p r\\right)+r(p q-9 r) \\geq 0\n\\end{aligned}\n\)\n\nThe last inequality follows from $N_{1}, N_{2}$ and $N_{3}$, and the fact that $p, q, r>0$. Equality occurs if and only if $x=y=z$.\n\nEquality holds when $x = y = z$, in which case $(xy + yz + zx) = 3x^2$ and $\frac{1}{(x+y)^2} + \frac{1}{(y+z)^2} + \frac{1}{(z+x)^2} = \frac{3}{(2x)^2} = \frac{3}{4x^2}$, so the left side is $3x^2 \cdot \frac{3}{4x^2} = \frac{9}{4}$. Thus, the minimum value of the expression is $C = \frac{9}{4}$.\n\nTherefore, the answer is $C = \frac{9}{4}$.

P269. Let $a, b, c$ be the lengths of the sides of a triangle. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$: \((a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c} \leq C (a^{a} b^{b} c^{c}).\)

S269. $C = 1$

By the weighted power mean inequality we have\n\(\n\\begin{aligned}\n& \\sqrt[a+b+c]{\\left(\\frac{a+b-c}{a}\\right)^{a}\\left(\\frac{b+c-a}{b}\\right)^{b}\\left(\\frac{c+a-b}{c}\\right)^{c}} \\\\\n& \\leq \\frac{1}{a+b+c}\\left(a \\cdot \\frac{a+b-c}{a}+b \\cdot \\frac{b+c-a}{b}+c \\cdot \\frac{c+a-b}{c}\\right)=1\n\\end{aligned}\n\)\ni.e.\n\(\n(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c} \\leq a^{a} b^{b} c^{c}\n\)\n\nEquality occurs when $a = b = c$, that is, for an equilateral triangle. In this case, both sides of the inequality are equal, so $C = 1$ is the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P270. Let $x, y, z \in \mathbb{R}^{+}$ such that $x y z = 1$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq x + y + z$. Determine the largest constant $C$ such that for any natural number $n$, the following inequality holds for all $x, y, z$: \(\frac{1}{x^{n}} + \frac{1}{y^{n}} + \frac{1}{z^{n}} \geq C (x^{n} + y^{n} + z^{n}).\)

S270. $C = 1$

After setting $x=\frac{a}{b}, y=\frac{b}{c}$ and $z=\frac{c}{a}$, the initial condition\n\(\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\geq x+y+z\n\)\nbecomes\n\(\n\\begin{aligned}\n\\frac{b}{a} & +\\frac{c}{b}+\\frac{a}{c} \\geq \\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a} \\\\\n& \\Leftrightarrow \\quad a^{2} b+b^{2} c+c^{2} a \\geq a b^{2}+b c^{2}+c a^{2} \\\\\n& \\Leftrightarrow \\quad(a-b)(b-c)(c-a) \\leq 0\n\\end{aligned}\n\)\n\nLet $n \in \mathbb{N}$, and take $A=a^{n}, B=b^{n}, C=c^{n}$.\nThen $a \geq b \Leftrightarrow A \geq B$ and $a \leq b \Leftrightarrow A \leq B$, etc.\nSo we have\n\(\n\\begin{aligned}\n&(A-B)(B-C)(C-A) \\leq 0 \\\\\n& \\Leftrightarrow \\quad \\frac{B}{A}+\\frac{C}{B}+\\frac{A}{C} \\geq \\frac{A}{B}+\\frac{B}{C}+\\frac{C}{A} \\\\\n& \\Leftrightarrow \\quad \\frac{1}{x^{n}}+\\frac{1}{y^{n}}+\\frac{1}{z^{n}} \\geq x^{n}+y^{n}+z^{n}\n\\end{aligned}\n\)\nExpected Answer: $C = 1$\n\nEquality holds when $a = b = c$, which corresponds to $x = y = z = 1$. In this case, $\frac{1}{x^{n}} + \frac{1}{y^{n}} + \frac{1}{z^{n}} = 3$ and $x^{n} + y^{n} + z^{n} = 3$, so the inequality becomes $3 \geq C \cdot 3$, i.e., $C = 1$. This gives the minimum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P271. Let $x, y, z \in \mathbb{R}^{+}$ such that $x + y + z = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $x, y, z$: \(\frac{1}{1-x y}+\frac{1}{1-y z}+\frac{1}{1-z x} \leq C.\)

S271. $C = \frac{27}{8}$

Let $p=x+y+z=1, q=x y+y z+z x, r=x y z$.\nIt can easily be shown that\n\(\n(1-x y)(1-y z)(1-z x)=1-q+p r-r^{2}\n\)\nand\n\(\n(1-x y)(1-y z)+(1-y z)(1-z x)+(1-z x)(1-x y)=3-2 q+p r\n\)\n\nSo the given inequality becomes\n\(\n\\begin{gathered}\n8(3-2 q+p r) \\leq 27\\left(1-q+p r-r^{2}\\right) \\\\\n\\Leftrightarrow \\quad 3-11 q+19 p r-27 r^{2} \\geq 0\n\\end{gathered}\n\)\n\nSince $p=1$, we need to show that\n\(\n3-11 q+19 r-27 r^{2} \\geq 0\n\)\n\nBy $N_{5}: p^{3} \geq 27 r$ we have $1 \geq 27 r$, i.e. $r \geq 27 r^{2}$.\n\nTherefore\n\(\n3-11 q+19 r-27 r^{2} \\geq 3-11 q+19 r-r=3-11 q+18 r\n\)\n\nSo it suffices to prove that\n\(\n3-11 q+18 r \\geq 0\n\)\n\nWe have\n\(\n\\begin{aligned}\n& 3-11 q+18 r \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad 3-11(x y+y z+z x)+18 x y z \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad 11(x y+y z+z x)-18 x y z \\leq 3\n\\end{aligned}\n\)\n\nApplying $A M \geq G M$ we deduce\n\(\n\\begin{aligned}\n11(x y+y z+z x)-18 x y z & =x y(11-18 z)+11 z(x+y) \\\\\n& \\leq \\frac{(x+y)^{2}}{4}(11-18 z)+11 z(x+y) \\\\\n& =(1-z)^{2}/4(11-18 z)+11 z(1-z) \\\\\n& =\\frac{(1-z)((1-z)(11-18 z)+44 z)}{4} \\\\\n& =\\frac{4 z+3 z^{2}-18 z^{3}+11}{4}\n\\end{aligned}\n\)\n\nSo it remains to show that\n\(\n\\begin{aligned}\n& \\frac{4 z+3 z^{2}-18 z^{3}+11}{4} \\leq 3 \\\\\n& \\quad \\Leftrightarrow \\quad 4 z+3 z^{2}-18 z^{3} \\leq 1 \\\\\n& \\Leftrightarrow \\quad 18 z^{3}-3 z^{2}-4 z+1 \\geq 0 \\\\\n& \\Leftrightarrow \\quad(3 z-1)^{2}(2 z+1) \\geq 0\n\\end{aligned}\n\)\nwhich is obvious.\n\nEquality holds when $x = y = z = \frac{1}{3}$, since then $x + y + z = 1$ and $x y = y z = z x = \frac{1}{9}$, so $\frac{1}{1-x y} = \frac{1}{1-1/9} = \frac{9}{8}$, and the sum is $3 \times \frac{9}{8} = \frac{27}{8}$. This gives the maximum value of the sum, so $C = \frac{27}{8}$ is the minimal constant such that the inequality always holds.\n\nTherefore, the answer is $C = \frac{27}{8}$.

P272. Let $\alpha_{i}>0, i=1,2, \ldots, n$, be real numbers such that $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=1$. Determine the maximal constant $C$ such that the following inequality holds for all $\alpha_{i}$: \(\alpha_{1}^{\alpha_{1}} \alpha_{2}^{\alpha_{2}} \cdots \alpha_{n}^{\alpha_{n}} \geq C.\)

S272. $C = \frac{1}{n}$

If we take $a_{i}=\frac{1}{\alpha_{i}}, i=1,2, \ldots, n$, by the Weighted $AM-GM$ inequality we get\n\(\n\\frac{1}{\\alpha_{1}^{\\alpha_{1}}} \\frac{1}{\\alpha_{2}^{\\alpha_{2}}} \\cdots \\frac{1}{\\alpha_{n}^{\\alpha_{n}}} \\leq \\frac{1}{\\alpha_{1}} \\alpha_{1}+\\frac{1}{\\alpha_{2}} \\alpha_{2}+\\cdots+\\frac{1}{\\alpha_{n}} \\alpha_{n}=n\n\)\ni.e.\n\(\n\\frac{1}{n} \\leq \\alpha_{1}^{\\alpha_{1}} \\alpha_{2}^{\\alpha_{2}} \\cdots \\alpha_{n}^{\\alpha_{n}}\n\)\nEquality holds when $\alpha_1 = \alpha_2 = \cdots = \alpha_n = \frac{1}{n}$, in which case $\alpha_1^{\alpha_1} \cdots \alpha_n^{\alpha_n} = \left(\frac{1}{n}\right)^1 = \frac{1}{n}$. This gives the minimum value of the product, so $C = \frac{1}{n}$ is maximal.\n\nTherefore, the answer is $C = \frac{1}{n}$.

P273. Let $a, b \in \mathbb{R}^{+}$ and $n \in \mathbb{N}$. Find the smallest constant $C$ such that the following inequality holds for all $a, b \in \mathbb{R}^{+}$ and $n \in \mathbb{N}$: \((a+b)^{n}\left(a^{n}+b^{n}\right) \leq C\left(a^{2 n}+b^{2 n}\right)\)

S273. $C = 2^n$

By the power mean inequality, for any $x, y \in \mathbb{R}^{+}, n \in \mathbb{N}$, we have\n\(\n\\left(\\frac{x+y}{2}\\right)^{n} \\leq \\frac{x^{n}+y^{n}}{2}\n\)\n\nTherefore\n\(\n\\begin{aligned}\n(a+b)^{n}\\left(a^{n}+b^{n}\\right) & =2^{n}\\left(\\frac{a+b}{2}\\right)^{n}\\left(a^{n}+b^{n}\\right) \\\\\n& \\leq 2^{n}\\left(\\frac{a^{n}+b^{n}}{2}\\right)\\left(a^{n}+b^{n}\\right)=2^{n} \\frac{\\left(a^{n}+b^{n}\\right)^{2}}{2} \\\\\n& \\leq 2^{n} \\frac{2\\left(a^{2 n}+b^{2 n}\\right)}{2}=2^{n}\\left(a^{2 n}+b^{2 n}\\right)\n\\end{aligned}\n\)\n\nEquality holds when $a = b$, since all the inequalities above become equalities in this case. This gives the minimum value of $C$, so $C = 2^n$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2^n$.

P274. Let $a_{1}, a_{2}, \ldots, a_{n} \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$: \(\sum_{k=1}^{n} k a_{k} \leq C + \sum_{k=1}^{n} a_{k}^{k}\)

S274. $C = \binom{n}{2}$

For $1 \leq k \leq n$ we have\n\(\na_{k}^{k}+(k-1)=a_{k}^{k}+\\underbrace{1+1+\\cdots+1}_{k-1} \\geq k \\sqrt[k]{a_{k}^{k}}=k a_{k}\n\)\n\nAfter adding these inequalities, for $1 \leq k \leq n$ we get\n\(\n\\sum_{k=1}^{n} k a_{k} \\leq \\sum_{k=1}^{n} a_{k}^{k}+\\sum_{k=1}^{n}(k-1)=\\sum_{k=1}^{n} a_{k}^{k}+\\frac{n(n-1)}{2}=\\sum_{k=1}^{n} a_{k}^{k}+\\binom{n}{2}\n\)\n\nEquality holds when $a_k = 1$ for all $k$, since then $a_k^k = 1$ and $k a_k = k$, so both sides are equal. This gives the minimum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = \binom{n}{2}$.

P275. Let $a, b, c$ be positive real numbers such that $a + b + c \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(a + b + c \geq \frac{C}{a+b+c} + \frac{2}{abc}.\)

S275. $C = 3$

By $A M \geq H M$ we get\n\(\na+b+c \\geq \\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq \\frac{9}{a+b+c}\n\)\ni.e.\n\(\n\\begin{equation*}\n\\frac{a+b+c}{3} \\geq \\frac{3}{a+b+c} \\tag{1}\n\\end{equation*}\n\)\n\nWe will prove that\n\(\n\\begin{equation*}\n\\frac{2(a+b+c)}{3} \\geq \\frac{2}{a b c} \\tag{2}\n\\end{equation*}\n\)\ni.e.\n\(\na+b+c \\geq \\frac{3}{a b c}\n\)\n\nUsing the well-known inequality $(x y+y z+z x)^{2} \geq 3(x y+y z+z x)$ we obtain\n\(\n(a+b+c)^{2} \\geq\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)^{2} \\geq 3\\left(\\frac{1}{a b}+\\frac{1}{b c}+\\frac{1}{c a}\\right)=3 \\frac{a+b+c}{a b c}\n\)\ni.e.\n\(\na+b+c \\geq \\frac{3}{a b c}\n\)\n\nAfter adding (1) and (2) we get the required inequality.\n\nEquality holds when $a = b = c = 1$, since then $a + b + c = 3$, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 3$, and $abc = 1$, so the original constraint is satisfied with equality, and the inequality becomes $3 \geq \frac{3}{3} + 2 = 3$. This gives the maximum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = 3$.

P276. Let $a, b, c, x, y, z$ be positive real numbers such that $x + y + z = 1$. Determine the maximal constant $C$ such that the following inequality holds for all $a, b, c, x, y, z$:

S276. $C = 2$

The inequality being homogeneous in $a, b, c$ we can assume that $a+b+c=1$. We apply this time the AM-GM Inequality and we find that\n$a x+b y+c z+2 \sqrt{(x y+y z+z x)(a b+b c+c a)} \leq a x+b y+c z+x y+y z+z x+a b+b c+c a$.\nConsequently,\n$x y+y z+z x+a b+b c+c a=\frac{1-x^{2}-y^{2}-z^{2}}{2}+\frac{1-a^{2}-b^{2}-c^{2}}{2} \leq 1-a x-b y-c z$,\nthe last one being equivalent to $(x-a)^{2}+(y-b)^{2}+(z-c)^{2} \geq 0$.\n\nEquality holds when $x = a$, $y = b$, $z = c$, that is, when the vectors $(x, y, z)$ and $(a, b, c)$ are equal. In this case, the inequality becomes an equality, and $C = 2$ gives the maximal value for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.”, “We will use the Cauchy-Schwarz Inequality twice. First, we can write $a x+b y+$ $c z \leq \sqrt{a^{2}+b^{2}+c^{2}} \cdot \sqrt{x^{2}+y^{2}+z^{2}}$ and then we apply again the Cauchy-Schwarz Inequality to obtain:\n\n\(\n\\begin{aligned}\na x+b y+c z & +2 \\sqrt{(x y+y z+z x)(a b+b c+c a)} \\leq \\\\\n& \\leq \\sqrt{\\sum a^{2}} \\cdot \\sqrt{\\sum x^{2}}+\\sqrt{2 \\sum a b} \\cdot \\sqrt{2 \\sum x y} \\leq \\\\\n& \\leq \\sqrt{\\sum x^{2}+2 \\sum x y} \\cdot \\sqrt{\\sum a^{2}+2 \\sum a b}=\\sum a\n\\end{aligned}\n\)\n\nEquality in the above inequalities holds when $a = b = c$ and $x = y = z$, that is, when all variables are equal. In this case, the inequality becomes an equality, and $C = 2$ gives the maximal value for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.

P277. Let $a, b, c$ be positive real numbers such that $ab + bc + ca = 3$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(\left(a^{3}-a+5\right)\left(b^{5}-b^{3}+5\right)\left(c^{7}-c^{5}+5\right) \geq C\)

S277. $C = 125$

For any real number $x$, the numbers $x-1, x^{2}-1, x^{3}-1$ and $x^{5}-1$ are of the same sign.\n\nTherefore\n\(\n(x-1)\\left(x^{2}-1\\right) \\geq 0, \\quad\\left(x^{2}-1\\right)\\left(x^{3}-1\\right) \\geq 0 \\quad \\text { and } \\quad\\left(x^{2}-1\\right)\\left(x^{5}-1\\right) \\geq 0\n\)\ni.e.\n\(\n\\begin{aligned}\na^{3}-a^{2}-a+1 & \\geq 0 \\\\\nb^{5}-b^{3}-b^{2}+1 & \\geq 0 \\\\\nc^{7}-c^{5}-c^{2}+1 & \\geq 0\n\\end{aligned}\n\)\n\nSo it follows that\n\(\na^{3}-a+5 \\geq a^{2}+4, \\quad b^{5}-b^{3}+5 \\geq b^{2}+4 \\quad \\text { and } \\quad c^{7}-c^{5}+5 \\geq c^{2}+4\n\)\n\nMultiplying these inequalities gives us\n\(\n\\begin{equation*}\n\\left(a^{3}-a+5\\right)\\left(b^{5}-b^{3}+5\\right)\\left(c^{7}-c^{5}+5\\right) \\geq\\left(a^{2}+4\\right)\\left(b^{2}+4\\right)\\left(c^{2}+4\\right) \\tag{1}\n\\end{equation*}\n\)\n\nWe’ll prove that\n\(\n\\begin{equation*}\n\\left(a^{2}+4\\right)\\left(b^{2}+4\\right)\\left(c^{2}+4\\right) \\geq 25(a b+b c+c a+2) \\tag{2}\n\\end{equation*}\n\)\n\nWe have\n\(\n\\begin{aligned}\n& \\left(a^{2}+4\\right)\\left(b^{2}+4\\right)\\left(c^{2}+4\\right) \\\\\n& \\quad=a^{2} b^{2} c^{2}+4\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\\right)+16\\left(a^{2}+b^{2}+c^{2}\\right)+64\n\\end{aligned}\n\)\n\(\n\\begin{align*}\n= & a^{2} b^{2} c^{2}+\\left(a^{2}+b^{2}+c^{2}\\right)+2+4\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+3\\right) \\\\\n& +15\\left(a^{2}+b^{2}+c^{2}\\right)+50 \\tag{3}\n\\end{align*}\n\)\n\nBy the obvious inequalities\n$(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0 \quad$ and $\quad(a b-1)^{2}+(b c-1)^{2}+(c a-1)^{2} \geq 0$\nwe obtain\n\(\n\\begin{align*}\na^{2}+b^{2}+c^{2} & \\geq a b+b c+c a \\tag{4}\\\\\na^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+3 & \\geq 2(a b+b c+c a) \\tag{5}\n\\end{align*}\n\)\n\nWe’ll prove that\n\(\n\\begin{equation*}\na^{2} b^{2} c^{2}+\\left(a^{2}+b^{2}+c^{2}\\right)+2 \\geq 2(a b+b c+c a) \\tag{6}\n\\end{equation*}\n\)\n\nLemma Let $x, y, z>0$. Then\n\(\n3 x y z+x^{3}+y^{3}+z^{3} \\geq 2\\left((x y)^{3 / 2}+(y z)^{3 / 2}+(z x)^{3 / 2}\\right)\n\)\n\nProof By Schur’s inequality and $A M \geq G M$ we have\n\(\n\\begin{aligned}\nx^{3}+y^{3}+z^{3}+3 x y z & \\geq\\left(x^{2} y+y^{2} x\\right)+\\left(z^{2} y+y^{2} z\\right)+\\left(x^{2} z+z^{2} x\\right) \\\\\n& \\geq 2\\left((x y)^{3 / 2}+(y z)^{3 / 2}+(z x)^{3 / 2}\\right)\n\\end{aligned}\n\)\n\nBy Lemma for $x=a^{2 / 3}, y=b^{2 / 3}, z=c^{2 / 3}$ we deduce\n\(\n3(a b c)^{2 / 3}+a^{2}+b^{2}+c^{2} \\geq 2(a b+b c+c a)\n\)\n\nTherefore it suffices to prove that\n\(\na^{2} b^{2} c^{2}+2 \\geq 3(a b c)^{2 / 3}\n\)\nwhich follows immediately by $A M \geq G M$.\nThus we have proved inequality (6).\nNow by (3), (4), (5) and (6) we obtain inequality (2).\nFinally by (1), (2) and since $a b+b c+c a=3$ we obtain the required inequality. Equality occurs if and only if $a=b=c=1$.\n\nEquality holds when $a = b = c = 1$, which satisfies $ab + bc + ca = 3$. In this case, the expression evaluates to $(1^3 - 1 + 5)(1^5 - 1^3 + 5)(1^7 - 1^5 + 5) = (5)(5)(5) = 125$. This gives the minimum value of the expression, so $C = 125$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 125$.

P278. Let $n$ be an integer greater than 2. Find the greatest real number $C_{\text{min}}$ and the least real number $C_{\text{max}}$ such that for any positive real numbers $x_1, x_2, \ldots, x_n$ (with $x_n = x_0$ and $x_{n+1} = x_1$), the following inequality holds: \(C_{\text{min}} \leq \sum_{i=1}^{n} \frac{x_i}{x_{i-1} + 2(n-1)x_i + x_{i+1}} \leq C_{\text{max}}\)

S278. $C_{\text{min}} = \frac{1}{2(n-1)}$ $C_{\text{max}} = \frac{1}{2}$

We will prove that $m_{n}=\frac{1}{2(n-1)}, M_{n}=\frac{1}{2}$. First, let us see that the inequality\n\n\(\n\\sum_{i=1}^{n} \\frac{x_{i}}{x_{i-1}+2(n-1) x_{i}+x_{i+1}} \\geq \\frac{1}{2(n-1)}\n\)\n\nis trivial, because $x_{i-1}+2(n-1) x_{i}+x_{i+1} \leq 2(n-1) \cdot \sum_{k=1}^{n} x_{k}$ for all $i$. This shows that $m_{n} \geq \frac{1}{2(n-1)}$. Taking $x_{i}=x^{i}$, the expression becomes\n\n\(\n\\frac{1}{x+x^{n-1}+2(n-1)}+\\frac{(n-2) x}{1+2(n-1) x+x^{2}}+\\frac{x^{n-1}}{1+2(n-1) x^{n-1}+x^{n-2}}\n\)\n\nand taking the limit when $x$ approaches 0 , we find that $m_{n} \leq \frac{1}{2(n-1)}$ and thus $m_{n}=\frac{1}{2(n-1)}$\n\nNow, we will prove that $M_{n} \geq \frac{1}{2}$. Of course, it suffices to prove that for any $x_{1}, x_{2}, \ldots, x_{n}>0$ we have\n\n\(\n\\sum_{i=1}^{n} \\frac{x_{i}}{x_{i-1}+2(n-1) x_{i}+x_{i+1}} \\leq \\frac{1}{2}\n\)\n\nBut it is clear that\n\n\(\n\\begin{gathered}\n\\sum_{i=1}^{n} \\frac{2 x_{i}}{x_{i-1}+2(n-1) x_{i}+x_{i+1}} \\leq \\sum_{i=1}^{n} \\frac{2 x_{i}}{2 \\sqrt{x_{i-1} \\cdot x_{i+1}}+2(n-1) x_{i}}= \\\\\n=\\sum_{i=1}^{n} \\frac{1}{n-1+\\frac{\\sqrt{x_{i-1} \\cdot x_{i+1}}}{x_{i}}}\n\\end{gathered}\n\)\n\nTaking $\frac{\sqrt{x_{i-1} \cdot x_{i+1}}}{x_{i}}=a_{i}$, we have to prove that if $\prod_{i=1}^{n} a_{i}=1$ then $\sum_{i=1}^{n} \frac{1}{n-1+a_{i}} \leq 1$. But this has already been proved in the problem 84. Thus, $M_{n} \geq \frac{1}{2}$ and because for $x_{1}=x_{2}=\cdots=x_{n}$ we have equality, we deduce that $M_{n}=\frac{1}{2}$, which solves the problem.\n\nEquality in the lower bound $C_{\text{min}} = \frac{1}{2(n-1)}$ is achieved in the limit as one variable tends to zero and the others are fixed, for example, by taking $x_1 = x_2 = \cdots = x_{n-1} = 1$ and $x_n \to 0$. Equality in the upper bound $C_{\text{max}} = \frac{1}{2}$ is achieved when all variables are equal, i.e., $x_1 = x_2 = \cdots = x_n$. Thus, $C_{\text{min}}$ is the minimum value and $C_{\text{max}}$ is the maximum value of the given sum.\n\nTherefore, the answer is \(C_{\\text{min}} = \\frac{1}{2(n-1)}\)\n\(C_{\\text{max}} = \\frac{1}{2}\).

P279. Let $x_{1}, x_{2}, \ldots, x_{n}$ be non-negative real numbers such that $x_{1} + x_{2} + \cdots + x_{n} \leq \frac{1}{2}$. Find the largest constant $C$ such that the following inequality holds for all $x_{1}, x_{2}, \ldots, x_{n}$ satisfying the given constraint: \(\left(1 - x_{1}\right)\left(1 - x_{2}\right) \cdots \left(1 - x_{n}\right) \geq C\)

S279. $C = \frac{1}{2}$

From $x_{1}+x_{2}+\cdots+x_{n} \leq \frac{1}{2}$ and the fact that $x_{1}, x_{2}, \ldots, x_{n}$ are nonnegative we deduce that\n\(\n0 \\leq x_{i} \\leq \\frac{1}{2}<1, \\quad \\text { i.e. } \\quad-x_{i}>-1, \\quad \\text { for all } i=1,2, \\ldots, n\n\)\nand it’s clear that all $-x_{i}$ are of the same sign.\nApplying Bernoulli’s inequality we obtain\n\(\n\\begin{aligned}\n\\left(1-x_{1}\\right)\\left(1-x_{2}\\right) \\cdots\\left(1-x_{n}\\right) & =\\left(1+\\left(-x_{1}\\right)\\right)\\left(1+\\left(-x_{2}\\right)\\right) \\cdots\\left(1+\\left(-x_{n}\\right)\\right) \\\\\n& \\geq 1+\\left(-x_{1}-x_{2}-\\cdots-x_{n}\\right) \\\\\n& =1-\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right) \\geq 1-\\frac{1}{2}=\\frac{1}{2}\n\\end{aligned}\n\)\nEquality holds when $x_{1} + x_{2} + \cdots + x_{n} = \frac{1}{2}$ and all $x_i \geq 0$, for example when $x_1 = \frac{1}{2}$ and $x_2 = x_3 = \cdots = x_n = 0$. This gives the minimum value of the product, so $C = \frac{1}{2}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{1}{2}$.

P280. Let $a, b, c \in (-3, 3)$ such that $\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c}=\frac{1}{3-a}+\frac{1}{3-b}+\frac{1}{3-c}$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given condition: \(\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c} \geq C\)

S280. $C = 1$

By the inequality $A M \geq H M$ we have\n\(\n\\begin{equation*}\n((3+a)+(3+b)+(3+c))\\left(\\frac{1}{3+a}+\\frac{1}{3+b}+\\frac{1}{3+c}\\right) \\geq 9 \\tag{1}\n\\end{equation*}\n\)\nand\n\(\n\\begin{align*}\n& ((3-a)+(3-b)+(3-c))\\left(\\frac{1}{3-a}+\\frac{1}{3-b}+\\frac{1}{3-c}\\right) \\geq 9 \\\\\n& \\quad \\Leftrightarrow \\quad((3-a)+(3-b)+(3-c))\\left(\\frac{1}{3+a}+\\frac{1}{3+b}+\\frac{1}{3+c}\\right) \\geq 9 \\tag{2}\n\\end{align*}\n\)\n\nAfter adding (1) and (2) we obtain\n\(\n18\\left(\\frac{1}{3+a}+\\frac{1}{3+b}+\\frac{1}{3+c}\\right) \\geq 18, \\quad \\text { i.e. } \\quad \\frac{1}{3+a}+\\frac{1}{3+b}+\\frac{1}{3+c} \\geq 1\n\)\n\nEquality holds when $a = b = c = 0$, since then $\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c} = 3 \times \frac{1}{3} = 1$. This gives the minimum value of the sum, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P281. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(\frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq C\)

S281. $C = 1$

Applying $A M \geq G M$ gives us\n\(\n\\frac{a^{2}}{a+2 b^{3}}=a-\\frac{2 a b^{3}}{a+2 b^{3}} \\geq a-\\frac{2 a b^{3}}{3 \\sqrt[3]{a b^{4}}}=a-\\frac{2 b a^{2 / 3}}{3}\n\)\n\nAnalogously\n\(\n\\frac{b^{2}}{b+2 c^{3}} \\geq b-\\frac{2 c b^{2 / 3}}{3} \\quad \\text { and } \\quad \\frac{c^{2}}{c+2 a^{3}} \\geq c-\\frac{2 a c^{2 / 3}}{3}\n\)\n\nAdding these three inequalities implies\n\(\n\\frac{a^{2}}{a+2 b^{2}}+\\frac{b^{2}}{b+2 c^{2}}+\\frac{c^{2}}{c+2 a^{2}} \\geq(a+b+c)-\\frac{2}{3}\\left(b a^{2 / 3}+c b^{2 / 3}+a c^{2 / 3}\\right)\n\)\n\nSo it is enough to prove that\n\(\n(a+b+c)-\\frac{2}{3}\\left(b a^{2 / 3}+c b^{2 / 3}+a c^{2 / 3}\\right) \\geq 1\n\)\ni.e.\n\(\n\\begin{equation*}\nb a^{2 / 3}+c b^{2 / 3}+a c^{2 / 3} \\leq 3 \\tag{1}\n\\end{equation*}\n\)\n\nAfter another application of $A M \geq G M$ we get\n\(\n\\begin{aligned}\nb a^{2 / 3}+c b^{2 / 3}+a c^{2 / 3} & \\leq \\frac{b(2 a+1)+c(2 b+1)+a(2 c+1)}{3} \\\\\n& =\\frac{a+b+c+2(a b+b c+c a)}{3} \\\\\n& \\leq \\frac{(a+b+c)+(a+b+c)^{2} / 3}{3}=\\frac{3+2 \\cdot 3^{2} / 3}{3}=3\n\\end{aligned}\n\)\ni.e. we have proved (1), and we are done.\n\nEquality holds when $a = b = c = 1$, since then $a + b + c = 3$ and all terms are equal, so the minimum value of the sum is achieved and $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P282. Let $t_{a}, t_{b}, t_{c}$ be the lengths of the medians, and $a, b, c$ be the lengths of the sides of a given triangle. Find the smallest constant $C$ such that the following inequality holds for all triangles: \(t_{a} t_{b}+t_{b} t_{c}+t_{c} t_{a} < C(a b+b c+c a)\)

S282. $C = \frac{5}{4}$

We can easily show the inequalities\n\(\nt_{a}<\\frac{b+c}{2}, \\quad t_{b}<\\frac{a+c}{2}, \\quad t_{c}<\\frac{b+a}{2}\n\)\n\nAfter adding these we get\n\(\n\\begin{equation*}\nt_{a}+t_{b}+t_{c}<a+b+c \\tag{1}\n\\end{equation*}\n\)\n\nBy squaring (1) we deduce\n\(\n\\begin{equation*}\nt_{a}^{2}+t_{b}^{2}+t_{c}^{2}+2\\left(t_{a} t_{b}+t_{b} t_{c}+t_{c} t_{a}\\right)<a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \\tag{2}\n\\end{equation*}\n\)\n\nOn the other hand, we have\n\(\nt_{a}^{2}=\\frac{2\\left(b^{2}+c^{2}\\right)-a^{2}}{4}, \\quad t_{b}^{2}=\\frac{2\\left(a^{2}+c^{2}\\right)-b^{2}}{4}, \\quad t_{c}^{2}=\\frac{2\\left(b^{2}+a^{2}\\right)-c^{2}}{4}\n\)\nso\n\(\nt_{a}^{2}+t_{b}^{2}+t_{c}^{2}=\\frac{3}{4}\\left(a^{2}+b^{2}+c^{2}\\right)\n\)\n\nNow using the previous result and (2) we get\n\(\n\\begin{equation*}\nt_{a} t_{b}+t_{b} t_{c}+t_{c} t_{a}<\\frac{1}{8}\\left(a^{2}+b^{2}+c^{2}\\right)+(a b+b c+c a) \\tag{3}\n\\end{equation*}\n\)\n\nAlso we have $a^{2}+b^{2}+c^{2}<2(a b+b c+c a)$, since\n$a^{2}+b^{2}+c^{2}-2(a b+b c+c a)=a(a-b-c)+b(b-a-c)+c(c-a-b)<0$.\nFinally by (3) and the previous inequality we obtain\n\(\nt_{a} t_{b}+t_{b} t_{c}+t_{c} t_{a}<\\frac{5}{4}(a b+b c+c a)\n\)\n\nEquality in the above inequalities is never achieved for any non-degenerate triangle, since all the inequalities used are strict. Thus, the value $C = \frac{5}{4}$ is the smallest constant for which the inequality always holds, but the inequality is always strict and never achieved with equality for any triangle.\n\nTherefore, the answer is $C = \frac{5}{4}$.

P283. Let $a, b, c \in \mathbb{R}$ such that $a + b + c \geq abc$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(a^2 + b^2 + c^2 \geq C \, abc.\)

S283. $C = \sqrt{3}$

We have\n\(\n\\begin{align*}\n\\left(a^{2}+b^{2}+c^{2}\\right)^{2} & =a^{4}+b^{4}+c^{4}+2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2} \\\\\n& =a^{4}+b^{4}+c^{4}+a^{2}\\left(b^{2}+c^{2}\\right)+b^{2}\\left(c^{2}+a^{2}\\right)+c^{2}\\left(a^{2}+b^{2}\\right) \\tag{1}\n\\end{align*}\n\)\n\nBy Exercise 1.7, it follows that\n\(\n\\begin{equation*}\na^{4}+b^{4}+c^{4} \\geq a b c(a+b+c) \\tag{2}\n\\end{equation*}\n\)\n\nAlso\n\(\n\\begin{equation*}\nb^{2}+c^{2} \\geq 2 b c, \\quad c^{2}+a^{2} \\geq 2 c a, \\quad a^{2}+b^{2} \\geq 2 a b \\tag{3}\n\\end{equation*}\n\)\n\nNow by (1), (2) and (3) we deduce\n\(\n\\begin{align*}\n\\left(a^{2}+b^{2}+c^{2}\\right)^{2} & \\geq a b c(a+b+c)+2 a^{2} b c+2 b^{2} a c+2 c^{2} a b \\\\\n& =a b c(a+b+c)+2 a b c(a+b+c)=3 a b c(a+b+c) \\tag{4}\n\\end{align*}\n\)\n\nSince $a+b+c \geq a b c$ in (4) we have\n\(\n\\left(a^{2}+b^{2}+c^{2}\\right)^{2} \\geq 3 a b c(a+b+c) \\geq 3(a b c)^{2}\n\)\ni.e.\n\(\na^{2}+b^{2}+c^{2} \\geq \\sqrt{3} a b c .\n\)\n\nEquality holds when $a = b = c = \sqrt{3}$, since then $a + b + c = 3\sqrt{3}$ and $abc = (\sqrt{3})^3 = 3\sqrt{3}$, so the constraint is satisfied with equality, and $a^2 + b^2 + c^2 = 3 \times 3 = 9 = \sqrt{3} \times 3\sqrt{3} = 9$. This gives the minimum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = \sqrt{3}$.

P284. Let $a, b, c \in \mathbb{R}^{+}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: \(a^{4}+b^{4}+c^{4} \geq C \cdot a b c (a+b+c)\)

S284. $C = 1$

We have\n\(\n\\begin{aligned}\na^{4} & +b^{4}+c^{4} \\geq a b c(a+b+c) \\\\\n& \\Leftrightarrow \\quad a^{4}+b^{4}+c^{4} \\geq a^{2} b c+b^{2} a c+c^{2} a b \\\\\n& \\Leftrightarrow \\quad \\frac{T[4,0,0]}{2} \\geq \\frac{T[2,1,1]}{2}\n\\end{aligned}\n\)\ni.e.\n\(\nT[4,0,0] \\geq T[2,1,1]\n\)\nwhich is true according to Muirhead’s theorem.\n\nEquality holds when $a = b = c$, since then both sides are equal: $a^4 + b^4 + c^4 = 3a^4$ and $abc(a+b+c) = a^3 \cdot 3a = 3a^4$. This gives the minimum value of the left side, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P285. Let $a, b, c \in \mathbb{R}^{+}$ such that $ab + bc + ca = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(\frac{1}{a(a+b)}+\frac{1}{b(b+c)}+\frac{1}{c(c+a)} \geq C\)

S285. $C = \frac{9}{2}$

The given inequality is equivalent to\n\(\n\\frac{c(a+b)+a b}{a(a+b)}+\\frac{a(b+c)+b c}{b(b+c)}+\\frac{b(c+a)+a c}{c(c+a)} \\geq \\frac{9}{2}\n\)\ni.e.\n\(\n\\begin{align*}\n\\frac{a}{b} & +\\frac{b}{c}+\\frac{c}{a}+\\frac{b}{a+b}+\\frac{c}{b+c}+\\frac{a}{c+a} \\geq \\frac{9}{2} \\\\\n& \\Leftrightarrow \\quad \\frac{a+b}{b}+\\frac{b+c}{c}+\\frac{c+a}{a}+\\frac{b}{a+b}+\\frac{c}{b+c}+\\frac{a}{c+a} \\geq \\frac{15}{2} \\tag{1}\n\\end{align*}\n\)\n\nWe have\n\(\n\\begin{aligned}\n& \\frac{a+b}{b}+\\frac{b+c}{c}+\\frac{c+a}{a}+\\frac{b}{a+b}+\\frac{c}{b+c}+\\frac{a}{c+a} \\\\\n& \\quad=\\frac{a+b}{4 b}+\\frac{b+c}{4 c}+\\frac{c+a}{4 a}+\\frac{b}{a+b}+\\frac{c}{b+c}+\\frac{a}{c+a}\n\\end{aligned}\n\)\n\(\n\\begin{aligned}\n& +\\frac{3}{4}\\left(\\frac{a+b}{b}+\\frac{b+c}{c}+\\frac{c+a}{a}\\right) \\\\\n\\geq & 6 \\sqrt[6]{\\frac{a+b}{4 b} \\cdot \\frac{b+c}{4 c} \\cdot \\frac{c+a}{4 a} \\cdot \\frac{b}{a+b} \\cdot \\frac{c}{b+c} \\cdot \\frac{a}{c+a}}+\\frac{3}{4}\\left(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+3\\right) \\\\\n\\geq & 3+\\frac{3}{4}\\left(3 \\sqrt[3]{\\frac{a}{b} \\cdot \\frac{b}{c} \\cdot \\frac{c}{a}}+3\\right)=3+\\frac{18}{4}=\\frac{15}{2}\n\\end{aligned}\n\)\nas required.\n\nEquality holds when $a = b = c = \frac{1}{\sqrt{3}}$, since $ab + bc + ca = 3 \left(\frac{1}{\sqrt{3}}\right)^2 = 1$, and in this case, all the steps above become equalities. This gives the minimum value of the sum, so $C = \frac{9}{2}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{9}{2}$.

P286. Let $a, b, c$ be positive real numbers. Find the smallest positive number $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: \(\frac{a+\sqrt{a b}+\sqrt[3]{a b c}}{C} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}\)

S286. $C = 3$

Applying $A M \geq G M$ we get\n\(\n\\sqrt[3]{a b \\cdot \\frac{a+b}{2}} \\geq \\sqrt[3]{a b \\cdot \\sqrt{a b}}=\\sqrt{a b}\n\)\n\nSo\n\(\na+\\sqrt{a b}+\\sqrt[3]{a b c} \\leq a+\\sqrt[3]{a b \\cdot \\frac{a+b}{2}}+\\sqrt[3]{a b c}\n\)\n\nNow, it is enough to show that\n\(\na+\\sqrt[3]{a b \\cdot \\frac{a+b}{2}}+\\sqrt[3]{a b c} \\leq 3 \\sqrt[3]{a \\cdot \\frac{a+b}{2} \\cdot \\frac{a+b+c}{3}}\n\)\n\nAnother application of $A M \geq G M$ gives us\n$\sqrt[3]{1 \cdot \frac{2 a}{a+b} \cdot \frac{3 a}{a+b+c}} \leq \frac{1+\frac{2 a}{a+b}+\frac{3 a}{a+b+c}}{3}, \quad \sqrt[3]{1 \cdot 1 \cdot \frac{3 b}{a+b+c}} \leq \frac{2+\frac{3 b}{a+b+c}}{3}$ and\n\(\n\\sqrt[3]{1 \\cdot \\frac{2 b}{a+b} \\cdot \\frac{3 c}{a+b+c}} \\leq \\frac{1+\\frac{2 b}{a+b}+\\frac{3 c}{a+b+c}}{3}\n\)\n\nAdding, we obtain\n\(\n\\sqrt[3]{\\frac{2 a}{a+b} \\cdot \\frac{3 a}{a+b+c}}+\\sqrt[3]{\\frac{3 b}{a+b+c}}+\\sqrt[3]{\\frac{2 b}{a+b} \\cdot \\frac{3 c}{a+b+c}} \\leq 3\n\)\ni.e.\n\(\n\\sqrt[3]{\\frac{1}{a} \\cdot \\frac{2}{a+b} \\cdot \\frac{3}{a+b+c}}\\left(a+\\sqrt[3]{a b \\cdot \\frac{a+b}{2}}+\\sqrt[3]{a b c}\\right) \\leq 3\n\)\ni.e.\n\(\na+\\sqrt[3]{a b \\cdot \\frac{a+b}{2}}+\\sqrt[3]{a b c} \\leq 3 \\sqrt[3]{a \\cdot \\frac{a+b}{2} \\cdot \\frac{a+b+c}{3}}\n\)\n\nEquality holds when $a = b = c$, that is, when all variables are equal. In this case, both sides of the inequality are equal, and the minimum value of $C$ for which the inequality always holds is $C = 3$.\n\nTherefore, the answer is $C = 3$.

P287. Let $a, b, c, d, e$ be positive real numbers such that $a+b+c+d+e=5$. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c, d, e$ satisfying the given constraint: \(a b c + b c d + c d e + d e a + e a b \leq C.\)

S287. $C = 5$

Without loss of generality, we may assume that $e=\min \{a, b, c, d, e\}$.\nBy $AM \geq GM$, we have\n\(\n\\begin{aligned}\na b c+b c d+c d e+d e a+e a b & =e(a+c)(b+d)+b c(a+d-e) \\\\\n& \\leq e\\left(\\frac{a+c+b+d}{2}\\right)^{2}+\\left(\\frac{b+c+a+d-e}{3}\\right)^{3} \\\\\n& =\\frac{e(5-e)^{2}}{4}+\\frac{(5-2 e)^{3}}{27}\n\\end{aligned}\n\)\n\nSo it suffices to prove that\n\(\n\\frac{e(5-e)^{2}}{4}+\\frac{(5-2 e)^{3}}{27} \\leq 5\n\)\nwhich can be rewrite as $(e-1)^{2}(e+8) \geq 0$, which is obviously true.\nEquality holds if and only if $a = b = c = d = e = 1$, that is, when all variables are equal. In this case, $a b c + b c d + c d e + d e a + e a b = 5$. Thus, this gives the maximum value of the expression, so the minimal constant $C$ is $5$.\n\nTherefore, the answer is $C = 5$.

P288. Let $a, b, c$ be non-negative real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}_0$: \(a^{3}+b^{3}+c^{3}+a b c \geq C(a+b+c)^{3}\)

S288. $C = \frac{1}{7}$

We have\n\(\n\\begin{aligned}\n(a+b+c)^{3} & =a^{3}+b^{3}+c^{3}+3\\left(a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\\right)+6 a b c \\\\\n& =\\frac{T[3,0,0]}{2}+3 T[2,1,0]+T[1,1,1]\n\\end{aligned}\n\)\nand\n\(\na^{3}+b^{3}+c^{3}+a b c=\\frac{T[3,0,0]}{2}+\\frac{T[1,1,1]}{6}\n\)\n\nSo we need to prove that\n\(\n7\\left(\\frac{T[3,0,0]}{2}+\\frac{T[1,1,1]}{6}\\right) \\geq \\frac{T[3,0,0]}{2}+3 T[2,1,0]+T[1,1,1]\n\)\ni.e.\n\(\n3 T[3,0,0]+\\frac{T[1,1,1]}{6} \\geq 3 T[2,1,0]\n\)\nwhich is true according to $T[3,0,0] \geq T[2,1,0]$ and $T[1,1,1] \geq 0$ (Muirhead’s theorem).\n\nEquality holds when $a = b = c$, since then all symmetric sums are equal and the inequality becomes an equality. This gives the minimum value of $C$, so $C = \frac{1}{7}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{1}{7}$.

P289. Let $a, b, c, d > 1$ be real numbers. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c, d > 1$: \(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1} \leq C \sqrt{(a b+1)(c d+1)}.\)

S289. $C = 1$

We’ll prove that for every $x, y \in \mathbb{R}^{+}$ we have $\sqrt{x-1}+\sqrt{y-1} \leq \sqrt{x y}$.\nApplying the Cauchy-Schwarz inequality for $a_{1}=\sqrt{x-1}, a_{2}=1 ; b_{1}=1, b_{2}=\sqrt{y-1}$ gives us\n\(\n(\\sqrt{x-1}+\\sqrt{y-1})^{2} \\leq x y, \\quad \\text { i.e. } \\quad \\sqrt{x-1}+\\sqrt{y-1} \\leq \\sqrt{x y}\n\)\n\nNow we easily deduce that\n\(\n\\sqrt{a-1}+\\sqrt{b-1}+\\sqrt{c-1}+\\sqrt{d-1} \\leq \\sqrt{a b}+\\sqrt{c d} \\leq \\sqrt{(a b+1)(c d+1)}\n\)\n\nEquality holds when $a = b = c = d = 2$, since then $\sqrt{a-1} = 1$ and $\sqrt{a b + 1} = \sqrt{5}$, so both sides equal $4$ and $\sqrt{(2\cdot2+1)(2\cdot2+1)} = 5$. Thus, the minimum value of $C$ for which the inequality always holds is $C = 1$.\n\nTherefore, the answer is $C = 1$.

P290. Let $n > 2$ and let $x_{1}, x_{2}, \ldots, x_{n} > 0$ such that $\sum_{i=1}^{n} x_i = 1$. Find the maximal constant $C$ such that the following inequality holds for all $x_{i}$: \(\prod_{i=1}^{n}\left(1+\frac{1}{x_{i}}\right) \geq C \prod_{i=1}^{n}\left(\frac{n-x_{i}}{1-x_{i}}\right).\)

S290. $C =1$

The most natural idea is to use the fact that\n\n\(\n\\frac{n-x_{i}}{1-x_{i}}=1+\\frac{n-1}{x_{1}+x_{2}+\\cdots+x_{i-1}+x_{i+1}+\\cdots+x_{n}}\n\)\n\nThus, we have\n\n\(\n\\prod_{i=1}^{n}\\left(\\frac{n-x_{i}}{1-x_{i}}\\right) \\leq \\prod_{i=1}^{n}\\left(1+\\frac{1}{\\sqrt[n-1]{x_{1} x_{2} \\ldots x_{i-1} x_{i+1} \\ldots x_{n}}}\\right)\n\)\n\nand we have to prove the inequality\n\n\(\n\\prod_{i=1}^{n}\\left(1+\\frac{1}{x_{i}}\\right) \\geq \\prod_{i=1}^{n}\\left(1+\\frac{1}{\\sqrt[n-1]{x_{1} x_{2} \\ldots x_{i-1} x_{i+1} \\ldots x_{n}}}\\right)\n\)\n\nBut this one is not very hard, because it follows immediately by multiplying the inequalities\n\n\(\n\\prod_{j \\neq i}\\left(1+\\frac{1}{x_{j}}\\right) \\geq\\left(1+\\sqrt[n-1]{\\prod_{j \\neq i} \\frac{1}{x_{j}}}\\right)^{n-1}\n\)\n\nobtained from Huygens Inequality.,We will prove even more, that\n\n\(\n\\prod_{i=1}^{n}\\left(1+\\frac{1}{x_{i}}\\right) \\geq\\left(\\frac{n^{2}-1}{n}\\right)^{n} \\cdot \\prod_{i=1}^{n} \\frac{1}{1-x_{i}}\n\)\n\nIt is clear that this inequality is stronger than the initial one. First, let us prove that\n\n\(\n\\prod_{i=1}^{n} \\frac{1+x_{i}}{1-x_{i}} \\geq\\left(\\frac{n+1}{n-1}\\right)^{n}\n\)\n\nThis follows from Jensen’s Inequality for the convex function $f(x)=\ln (1+x)-$ $\ln (1-x)$. So, it suffices to prove that\n\n\(\n\\frac{\\left(\\frac{n+1}{n-1}\\right)^{n}}{\\prod_{i=1}^{n} x_{i}} \\cdot \\prod_{i=1}^{n}\\left(1-x_{i}\\right)^{2} \\geq\\left(\\frac{n^{2}-1}{n}\\right)^{n}\n\)\n\nBut a quick look shows that this is exactly the inequality proved in the solution of the problem 121.\n\nEquality holds when $x_1 = x_2 = \cdots = x_n = \frac{1}{n}$, since then all terms are equal and the AM-GM and Jensen inequalities used above become equalities. This gives the minimum value of $C$ for which the inequality always holds, so the maximal constant is $C = 1$.\n\nTherefore, the answer is $C = 1$.

P291. Let $a, b, c \geq 0$ such that $a + b + c = 1$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$: \(a^{2} + b^{2} + c^{2} \leq C(a^{3} + b^{3} + c^{3}) + 3abc\)

S291. $C = 2$

Let $p=a+b+c=1, q=a b+b c+c a, r=a b c$.\nThe given inequality is equivalent to\n\(\n1-2 q \\leq 2(1-3 q+3 r)+3 r \\quad \\Leftrightarrow \\quad 4 q \\leq 1+9 r\n\)\n\nEquality holds when two of the variables are zero and the third is $1$, i.e., $(a, b, c) = (1, 0, 0)$ and its permutations. In this case, $q = 0$ and $r = 0$, so $4q = 0$ and $1 + 9r = 1$, and the inequality becomes $0 \leq 1$, which is tight. This gives the minimum value of $C$.\n\nTherefore, the answer is $C = 2$.

P292. Let $a, b, c, d > 0$ be real numbers such that $a \leq 1$, $a+b \leq 5$, $a+b+c \leq 14$, and $a+b+c+d \leq 30$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c, d$ satisfying the given constraints: \(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \leq C\)

S292. $C = 10$

The function $f:(0,+\infty) \rightarrow(0,+\infty)$ defined by $f(x)=\sqrt{x}$ is concave on $(0,+\infty)$, so by Jensen’s inequality, for\n\(\nn=4, \\quad \\alpha_{1}=\\frac{1}{10}, \\quad \\alpha_{2}=\\frac{2}{10}, \\quad \\alpha_{3}=\\frac{3}{10}, \\quad \\alpha_{4}=\\frac{4}{10}\n\)\nwe get\n\(\n\\frac{1}{10} \\sqrt{a}+\\frac{2}{10} \\sqrt{\\frac{b}{4}}+\\frac{3}{10} \\sqrt{\\frac{c}{9}}+\\frac{4}{10} \\sqrt{\\frac{d}{16}} \\leq \\sqrt{\\frac{a}{10}+\\frac{b}{20}+\\frac{c}{30}+\\frac{d}{40}}\n\)\ni.e.\n\(\n\\begin{equation*}\n\\sqrt{a}+\\sqrt{b}+\\sqrt{c}+\\sqrt{d} \\leq 10 \\sqrt{\\frac{12 a+6 b+4 c+3 d}{120}} \\tag{1}\n\\end{equation*}\n\)\n\nOn the other hand, we have\n\(\n\\begin{aligned}\n& 12 a+6 b+4 c+3 d \\\\\n& \\quad=3(a+b+c+d)+(a+b+c)+2(a+b)+6 a \\\\\n& \\quad \\leq 3 \\cdot 30+14+2 \\cdot 5+6 \\cdot 1=120\n\\end{aligned}\n\)\n\nBy (1) and the last inequality we obtain the required result.\n\nEquality holds when $a = 1$, $b = 4$, $c = 9$, $d = 16$, since these values satisfy all the constraints with equality and yield $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} = 1+2+3+4 = 10$. This gives the maximum value of the sum, so $C = 10$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 10$.

P293. Prove that if $n \geq 2$ and $a_{1}, a_{2}, \ldots, a_{n}$ are real numbers with product 1, then find the largest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$: \(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}-n \geq C \cdot\frac{n}{n-1} \cdot \sqrt[n]{n-1}\left(a_{1}+a_{2}+\cdots+a_{n}-n\right)\)

S293. $C = 2$

We will prove the inequality by induction. For $n=2$ it is trivial. Now, suppose the inequality is true for $n-1$ numbers and let us prove it for $n$. First, it is easy to see that it is enough to prove it for $a_{1}, \ldots, a_{n}>0$ (otherwise we replace $a_{1}, a_{2}, \ldots, a_{n}$ with $ㅣa_{1}ㅣ,ㅣa_{2}ㅣ, \ldots,ㅣa_{n}ㅣ$, which have product 1. Yet, the right hand side increases). Now, let $a_{n}$ the maximum number among $a_{1}, a_{2}, \ldots, a_{n}$ and let $G$ the geometric mean of $a_{1}, a_{2}, \ldots, a_{n-1}$. First, we will prove that\n\n\(\n\\begin{gathered}\na_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2}-n-\\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{1}+a_{2}+\\cdots+a_{n}-n\\right) \\geq \\\\\n\\geq a_{n}^{2}+(n-1) G^{2}-n-\\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{n}+(n-1) G-n\\right)\n\\end{gathered}\n\)\n\nwhich is equivalent to\n\n\(\na_{1}^{2}+a_{2}^{2}+\\cdots+a_{n-1}^{2}-(n-1) \\sqrt[n-1]{a_{1}^{2} a_{2}^{2} \\ldots a_{n-1}^{2}} \\geq\n\)\n\n\(\n\\geq \\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{1}+a_{2}+\\cdots+a_{n-1}-(n-1) \\sqrt[n-1]{a_{1} a_{2} \\ldots a_{n-1}}\\right)\n\)\n\nBecause, $\sqrt[n-1]{a_{1} a_{2} \ldots a_{n-1}} \leq 1$ and $a_{1}+a_{2}+\cdots+a_{n-1}-(n-1) \sqrt[n-1]{a_{1} a_{2} \ldots a_{n-1}} \geq$ 0 , it is enough to prove the inequality\n\n\(\na_{1}^{2}+\\cdots+a_{n-1}^{2}-(n-1) G^{2} \\geq \\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1} \\cdot G \\cdot\\left(a_{1}+\\cdots+a_{n-1}-(n-1) G\\right)\n\)\n\nNow, we apply the inductive hypothesis for the numbers $\frac{a_{1}}{G}, \ldots, \frac{a_{n-1}}{G}$ which have product 1 and we infer that\n\n\(\n\\frac{a_{1}^{2}+\\cdots+a_{n-1}^{2}}{G^{2}}-n+1 \\geq \\frac{2(n-1)}{n-2} \\cdot \\sqrt[n-1]{n-2}\\left(\\frac{a_{1}+\\cdots+a_{n-1}}{G}-n+1\\right)\n\)\n\nand so it suffices to prove that\n\n\(\n\\frac{2(n-1)}{n-2} \\cdot \\sqrt[n-1]{n-2}\\left(a_{1}+\\cdots+a_{n-1}-(n-1) G\\right) \\geq \\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{1}+\\cdots+a_{n-1}-(n-1) G\\right)\n\)\n\nwhich is the same as $1+\frac{1}{n(n-2)} \geq \frac{\sqrt[n]{n-1}}{\sqrt[n-1]{n-2}}$. This becomes\n\n\(\n\\left(1+\\frac{1}{n(n-2)}\\right)^{n(n-1)} \\geq \\frac{(n-1)^{n-1}}{(n-2)^{n}}\n\)\n\nand it follows for $n>4$ from\n\n\(\n\\left(1+\\frac{1}{n(n-2)}\\right)^{n(n-1)}>2\n\)\n\nand\n\n\(\n\\frac{(n-1)^{n-1}}{(n-2)^{n}}=\\frac{1}{n-2} \\cdot\\left(1+\\frac{1}{n-2}\\right) \\cdot\\left(1+\\frac{1}{n-2}\\right)^{n-2}<\\frac{e}{n-2}\\left(1+\\frac{1}{n-2}\\right)<2\n\)\n\nFor $n=3$ and $n=4$ it is easy to check.\nThus, we have proved that\n\n\(\n\\begin{aligned}\na_{1}^{2} & +a_{2}^{2}+\\cdots+a_{n}^{2}-n-\\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{1}+a_{2}+\\cdots+a_{n}-n\\right) \\geq \\\\\n& \\geq a_{n}^{2}+(n-1) G^{2}-n-\\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{n}+(n-1) G-n\\right)\n\\end{aligned}\n\)\n\nand it is enough to prove that\n\n\(\nx^{2(n-1)}+\\frac{n-1}{x^{2}}-n \\geq \\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(x^{n-1}+\\frac{n-1}{x}-n\\right)\n\)\n\nfor all $x \geq 1$ (we took $x=\frac{1}{G}$ ). Let us consider the function\n\n\(\nf(x)=x^{2(n-1)}+\\frac{n-1}{x^{2}}-n-\\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(x^{n-1}+\\frac{n-1}{x}-n\\right)\n\)\n\nWe have\n\n\(\nf^{\\prime}(x)=2 \\cdot \\frac{x^{n}-1}{x^{2}} \\cdot\\left[\\frac{(n-1)\\left(x^{n}+1\\right)}{x}-n \\sqrt[n]{n-1}\\right] \\geq 0\n\)\n\nbecause\n\n\(\nx^{n-1}+\\frac{1}{x}=x^{n-1}+\\frac{1}{(n-1) x}+\\cdots+\\frac{1}{(n-1) x} \\geq n \\sqrt[n]{\\frac{1}{(n-1)^{n-1}}}\n\)\n\nThus, $f$ is increasing and so $f(x) \geq f(1)=0$. This proves the inequality.\n\nEquality holds when $a_1 = a_2 = \cdots = a_n = 1$, since then the product is 1, $a_1^2 + \cdots + a_n^2 = n$, and $a_1 + \cdots + a_n = n$, so both sides of the inequality are zero. This gives the minimum value of the left side, so $C = 2$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.

P294. Let $a, b, c$ be positive real numbers. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: \(\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}} \geq C \left( \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \right)\)

S294. $C = 1$

By the Cauchy-Schwarz Inequality we have\n\n\(\n(a+b+c)\\left(\\frac{a^{3}}{b^{2}}+\\frac{b^{3}}{c^{2}}+\\frac{c^{3}}{a^{2}}\\right) \\geq\\left(\\frac{a^{2}}{b}+\\frac{b^{2}}{c}+\\frac{c^{2}}{a}\\right)^{2}\n\)\n\nso we only have to prove that\n\n\(\n\\frac{a^{2}}{b}+\\frac{b^{2}}{c}+\\frac{c^{2}}{a} \\geq a+b+c\n\)\n\nBut this follows immediately from the Cauchy-Schwarz Inequality.\n\nEquality holds when $a = b = c$, in which case both sides of the original inequality are equal. This gives the minimum value of the left side relative to the right, so $C = 1$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$. “, “We have\n\n\(\n\\frac{a^{3}}{b^{2}} \\geq \\frac{a^{2}}{b}+a-b \\Leftrightarrow a^{3}+b^{3} \\geq a b(a+b) \\Leftrightarrow(a-b)^{2}(a+b) \\geq 0\n\)\n\nwhich is clearly true. Writing the analogous inequalities and adding them up gives\n\n\(\n\\frac{a^{3}}{b^{2}}+\\frac{b^{3}}{c^{2}}+\\frac{c^{3}}{a^{2}} \\geq \\frac{a^{2}}{b}+a-b+\\frac{b^{2}}{c}+b-c+\\frac{c^{2}}{a}+c-a=\\frac{a^{2}}{b}+\\frac{b^{2}}{c}+\\frac{c^{2}}{a}\n\)\n\nEquality holds when $a = b = c$, since then both sides are equal. This gives the minimum value of the left side relative to the right, so the maximal constant is $C = 1$.\n\nTherefore, the answer is $C = 1$.

P295. Let $a, b, c$ be positive real numbers such that $a b c \geq 1$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$: \(a+b+c \geq C\left(\frac{1+a}{1+b}+\frac{1+b}{1+c}+\frac{1+c}{1+a}\right).\)

S295. $C = 1$

We have\n\(\n\\begin{aligned}\na & +b+c-\\frac{1+a}{1+b}-\\frac{1+b}{1+c}-\\frac{1+c}{1+a} \\\\\n& =(1+a)\\left(1-\\frac{1}{1+b}\\right)+(1+b)\\left(1-\\frac{1}{1+c}\\right)+(1+c)\\left(1-\\frac{1}{1+a}\\right)-3 \\\\\n& =\\frac{(1+a) b}{1+b}+\\frac{(1+b) c}{1+c}+\\frac{(1+c) a}{1+a}-3 \\\\\n& \\geq 3 \\sqrt[3]{\\frac{(1+a) b}{1+b} \\cdot \\frac{(1+b) c}{1+c} \\cdot \\frac{(1+c) a}{1+a}}-3 \\\\\n& =3 \\sqrt[3]{a b c}-3 \\geq 0 \\quad(a b c \\geq 1)\n\\end{aligned}\n\)\n\nEquality holds when $a = b = c = 1$, since then $a + b + c = 3$ and $\frac{1+a}{1+b} + \frac{1+b}{1+c} + \frac{1+c}{1+a} = 3$, so the inequality becomes $3 \geq C \cdot 3$, which gives $C = 1$. This is the maximum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = 1$

P296. Let $h_{a}, h_{b}$, and $h_{c}$ be the lengths of the altitudes, and $R$ and $r$ be the circumradius and inradius, respectively, of a given triangle. Determine the minimal constant $C$ such that the following inequality holds for all triangles: \(h_{a}+h_{b}+h_{c} \leq 2R+Cr.\)

S296. $C = 5$

Lemma 21.4 In an arbitrary triangle we have\n\(\na b+b c+c a=r^{2}+s^{2}+4 r R \\quad \\text { and } \\quad a^{2}+b^{2}+c^{2}=2\\left(s^{2}-4 R r-r^{2}\\right)\n\)\n\nProof We have\n\(\n\\begin{aligned}\nr^{2}+s^{2}+4 r R & =\\frac{P^{2}}{s^{2}}+s^{2}+\\frac{a b c}{P} \\cdot \\frac{P}{s}=\\frac{(s-a)(s-b)(s-c)}{s}+s^{2}+\\frac{a b c}{s} \\\\\n& =\\frac{s^{3}-a s^{2}-b s^{2}-c s^{2}+a b s+b c s+c a s-a b c+s^{3}+a b c}{s} \\\\\n& =2 s^{2}-s(a+b+c)+a b+b c+c a \\\\\n& =2 s^{2}-2 s^{2}+a b+b c+c a=a b+b c+c a\n\\end{aligned}\n\)\n\nHence\n\(\n\\begin{equation*}\na b+b c+c a=r^{2}+s^{2}+4 r R \\tag{1}\n\\end{equation*}\n\)\n\nNow by (1) we have\n\(\n\\begin{aligned}\na b+b c+c a & =r^{2}+s^{2}+4 r R=\\frac{1}{2}\\left(2 r^{2}+8 r R+\\frac{(a+b+c)^{2}}{2}\\right) \\\\\n& =\\frac{1}{2}\\left(2 r^{2}+8 r R+\\frac{a^{2}+b^{2}+c^{2}}{2}\\right)+\\frac{a b+b c+c a}{2}\n\\end{aligned}\n\)\nfrom which it follows that\n\(\n\\begin{equation*}\na b+b c+c a=2 r^{2}+8 r R+\\frac{a^{2}+b^{2}+c^{2}}{2} \\tag{2}\n\\end{equation*}\n\)\n\nNow (1) and (2) yields\n\(\n\\begin{equation*}\na^{2}+b^{2}+c^{2}=2\\left(s^{2}-4 R r-r^{2}\\right) \\tag{3}\n\\end{equation*}\n\)\n\nWithout proof we will give the following lemma.\n\nIn an arbitrary triangle we have\n\(\n\\begin{equation*}\ns^{2} \\leq 4 R^{2}+4 R r+3 r^{2} \\tag{4}\n\\end{equation*}\n\)\n\nIn an arbitrary triangle we have $a^{2}+b^{2}+c^{2} \leq 8 R^{2}+4 r^{2}$.\nProof From (3) and (4) we have\n\(\na^{2}+b^{2}+c^{2}=2\\left(s^{2}-4 R r-r^{2}\\right) \\leq 2\\left(4 R^{2}+4 R r+3 r^{2}-4 R r-r^{2}\\right)=8 R^{2}+4 r^{2}\n\)\n\nHence\n\(\n\\begin{equation*}\na^{2}+b^{2}+c^{2} \\leq 8 R^{2}+4 r^{2} \\tag{5}\n\\end{equation*}\n\)\n\nNow let us consider our problem.\nWe have\n\(\n\\begin{aligned}\n& 2 R\\left(h_{a}+h_{b}+h_{c}\\right)=2 R\\left(\\frac{2 P}{a}+\\frac{2 P}{b}+\\frac{2 P}{c}\\right)=4 P R \\frac{a b+b c+c a}{a b c} \\\\\n&=a b+b c+c a \\\\\n& \\stackrel{(2)}{=} 2 r^{2}+8 r R+\\frac{a^{2}+b^{2}+c^{2}}{2} \\\\\n& \\stackrel{(4)}{\\leq} 2 r^{2}+8 r R+4 R^{2}+2 r^{2} \\\\\n& \\Leftrightarrow \\quad R\\left(h_{a}+h_{b}+h_{c}\\right) \\leq 2 R^{2}+4 R r+2 r^{2} \\leq 2 R^{2}+4 R r+R r \\leq R(2 R+5 r)\n\\end{aligned}\n\)\n\nHence\n\(\nh_{a}+h_{b}+h_{c} \\leq 2 R+5 r\n\)\n\nEquality occurs iff $a=b=c$.\n\nEquality holds when the triangle is equilateral, i.e., $a = b = c$. In this case, the bound is achieved, and $C = 5$ gives the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 5$.

P297. Let $a, b, c > 0$ be real numbers such that $a + b + c = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: \(\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b} \geq C\)

S297. $C = 2$

Applying the Cauchy-Schwarz inequality for the sequences\n\(\na_{1}=\\sqrt{\\frac{a^{2}+b}{b+c}}, \\quad a_{2}=\\sqrt{\\frac{b^{2}+c}{c+a}}, \\quad a_{3}=\\sqrt{\\frac{c^{2}+a}{a+b}}\n\)\nand\n$b_{1}=\sqrt{\left(a^{2}+b\right)(b+c)}, \quad b_{2}=\sqrt{\left(b^{2}+c\right)(c+a)}, \quad b_{3}=\sqrt{\left(c^{2}+a\right)(a+b)}$\nwe obtain\n$\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b} \geq \frac{\left(a^{2}+b^{2}+c^{2}+1\right)^{2}}{\left(a^{2}+b\right)(b+c)+\left(b^{2}+c\right)(c+a)+\left(c^{2}+a\right)(a+b)}$.\nSo it suffices to show that\n\(\n\\frac{\\left(a^{2}+b^{2}+c^{2}+1\\right)^{2}}{\\left(a^{2}+b\\right)(b+c)+\\left(b^{2}+c\\right)(c+a)+\\left(c^{2}+a\\right)(a+b)} \\geq 2\n\)\n\nWe have\n\(\n\\begin{aligned}\n& \\frac{\\left(a^{2}+b^{2}+c^{2}\\right.}{\\left(a^{2}+b\\right)(b+c)+\\left(b^{2}+c\\right)(c+a)+\\left(c^{2}+a\\right)(a+b)} \\geq 2 \\\\\n& \\Leftrightarrow \\quad\\left(a^{2}+b^{2}+c^{2}+1\\right)^{2} \\geq 2\\left(\\left(a^{2}+b\\right)(b+c)+\\left(b^{2}+c\\right)(c+a)\\right. \\\\\n&\\left.+\\left(c^{2}+a\\right)(a+b)\\right) \\\\\n& \\Leftrightarrow \\quad 1+\\left(a^{2}+b^{2}+c^{2}\\right)^{2} \\geq 2\\left(a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\\right) \\\\\n&+2(a b+b c+c a) \\\\\n& \\Leftrightarrow \\quad 1+\\left(a^{2}+b^{2}+c^{2}\\right)^{2} \\geq 2\\left(a^{2}(1-a)+b^{2}(1-b)+c^{2}(1-c)\\right) \\\\\n&+2(a b+b c+c a)\n\\end{aligned}\n\)\n\(\n\\begin{aligned}\n\\Leftrightarrow \\quad 1+\\left(a^{2}+b^{2}+c^{2}\\right)^{2} \\geq & 2\\left(a^{2}+b^{2}+c^{2}-a^{3}-b^{3}-c^{3}\\right) \\\\\n& +2(a b+b c+c a) \\\\\n\\Leftrightarrow \\quad\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+2\\left(a^{3}+b^{3}+c^{3}\\right) \\geq & 2\\left(a^{2}+b^{2}+c^{2}+a b+b c+c a\\right)-1 \\\\\n\\Leftrightarrow \\quad\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+2\\left(a^{3}+b^{3}+c^{3}\\right) \\geq & 2(a(1-c)+b(1-a) \\\\\n& +c(1-b))-1 \\\\\n\\Leftrightarrow \\quad\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+2\\left(a^{3}+b^{3}+c^{3}\\right) \\geq & 1-2(a b+b c+c a) \\\\\n\\Leftrightarrow \\quad\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+2\\left(a^{3}+b^{3}+c^{3}\\right) \\geq & \\geq(a+b+c)^{2}-2(a b+b c+c a) \\\\\n& =a^{2}+b^{2}+c^{2} .\n\\end{aligned}\n\)\n\nSo we need to show that\n\(\n\\begin{equation*}\n\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+2\\left(a^{3}+b^{3}+c^{3}\\right) \\geq a^{2}+b^{2}+c^{2} \\tag{1}\n\\end{equation*}\n\)\n\nBy Chebishev’s inequality we deduce\n$(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \leq 3\left(a^{3}+b^{3}+c^{3}\right), \quad$ i.e. $\quad a^{3}+b^{3}+c^{3} \geq \frac{a^{2}+b^{2}+c^{2}}{3}$,\nand clearly $\left(a^{2}+b^{2}+c^{2}\right)^{2} \geq \frac{a^{2}+b^{2}+c^{2}}{3}$.\nAdding these inequalities gives us inequality (1).2 Take $a+b+c=p=1, a b+b c+c a=q, a b c=r$ and use the method from Chap. 14.\n\nEquality holds when $a = b = c = \frac{1}{3}$, since then $a + b + c = 1$ and the expression evaluates to $2$. This gives the minimum value of the sum, so $C = 2$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.

P298. Let $a_{1}, a_{2}, \ldots, a_{n}>1$ be positive integers. Find the minimal constant $C$ such that at least one of the numbers $\sqrt[a_{1}]{a_{2}}, \sqrt[a_{2}]{a_{3}}, \ldots, \sqrt[a_{n}]{\sqrt{a_{n}}}, \sqrt[a_{n}]{a_{1}}$ is less than or equal to $C$ for all sequences of positive integers $a_1, a_2, \ldots, a_n > 1$: \(\min \left( \sqrt[a_{1}]{a_{2}}, \sqrt[a_{2}]{a_{3}}, \ldots, \sqrt[a_{n}]{\sqrt{a_{n}}}, \sqrt[a_{n}]{a_{1}} \right) \leq C\)

S298. $C = \sqrt[3]{3}$

Suppose we have $a_{i+1}^{\frac{1}{a_{i}}}>3^{\frac{1}{3}}$ for all $i$. First, we will prove that $n^{\frac{1}{n}} \leq 3^{\frac{1}{3}}$ for all natural number $n$. For $n=1,2,3,4$ it is clear. Suppose the inequality is true for $n>3$ and let us prove it for $n+1$. This follows from the fact that\n\n\(\n1+\\frac{1}{n} \\leq 1+\\frac{1}{4}<\\sqrt[3]{3} \\Rightarrow 3^{\\frac{n+1}{3}}=\\sqrt[3]{3} \\cdot 3^{\\frac{n}{3}} \\geq \\frac{n+1}{n} \\cdot n=n+1\n\)\n\nThus, using this observation, we find that $a_{i+1}^{\frac{1}{a_{i}}}>3^{\frac{1}{3}} \geq a_{i+1}^{\frac{1}{a_{i+1}}} \Rightarrow a_{i+1}>a_{i}$ for all $i$, which means that $a_{1}<a_{2}<\cdots<a_{n-1}<a_{n}<a_{1}$, contradiction.\n\nEquality holds when $a_i = 3$ and $a_{i+1} = 3$ for some $i$, so $\sqrt[3]{3} = 3^{1/3}$. This gives the minimal possible value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = \sqrt[3]{3}$.

P299. Let $a, b, c, d$ be positive real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c, d$: \(\frac{a^2 + b^2 + 3}{a + b} + \frac{b^2 + c^2 + 3}{b + c} + \frac{c^2 + d^2 + 3}{c + d} + \frac{d^2 + a^2 + 3}{d + a} \geq C.\)

S299. $C = 10$

Observe that for any real numbers $x, y$ we have\n\(\nx^{2}+x y+y^{2}=\\left(x+\\frac{y}{2}\\right)^{2}+\\frac{3 y^{2}}{4} \\geq 0\n\)\nequality achieves if and only if $x=y=0$.\nHence $(a-1)^{2}+(a-1)(b-1)+(b-1)^{2} \geq 0$, which is equivalent to\n\(\na^{2}+b^{2}+a b-3 a-3 b+3 \\geq 0\n\)\nfrom which we obtain\n\(\na^{2}+b^{2}+3 \\geq 3 a+3 b-a b\n\)\ni.e.\n\(\n\\frac{a^{2}+b^{2}+3}{a+b} \\geq 3-\\frac{a b}{a+b}\n\)\n\nBy $A M \geq G M$ we easily deduce that\n\(\n\\frac{a+b}{4} \\geq \\frac{a b}{a+b}\n\)\n\nTherefore by previous inequality we get\n\(\n\\frac{a^{2}+b^{2}+3}{a+b} \\geq 3-\\frac{a+b}{4}\n\)\n\nSimilarly we obtain\n\(\n\\begin{aligned}\n& \\frac{b^{2}+c^{2}+3}{b+c} \\geq 3-\\frac{b+c}{4}, \\quad \\frac{c^{2}+d^{2}+3}{c+d} \\geq 3-\\frac{c+d}{4} \\quad \\text { and } \\\\\n& \\frac{d^{2}+a^{2}+3}{d+a} \\geq 3-\\frac{d+a}{4}\n\\end{aligned}\n\)\n\nAdding the last four inequality yields\n\(\n\\begin{equation*}\n\\frac{a^{2}+b^{2}+3}{a+b}+\\frac{b^{2}+c^{2}+3}{b+c}+\\frac{c^{2}+d^{2}+3}{c+d}+\\frac{d^{2}+a^{2}+3}{d+a} \\geq 12-\\frac{a+b+c+d}{2} \\tag{1}\n\\end{equation*}\n\)\n\nAccording to inequality $Q M \geq A M$ we deduce that\n\(\n\\sqrt{\\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}} \\geq \\frac{a+b+c+d}{4}\n\)\nand since $a^{2}+b^{2}+c^{2}+d^{2}=4$ we obtain\n\(\n\\begin{equation*}\na+b+c+d \\leq 4 \\tag{2}\n\\end{equation*}\n\)\n\nBy (1) and (2) we get\n\(\n\\begin{aligned}\n\\frac{a^{2}+b^{2}+3}{a+b}+\\frac{b^{2}+c^{2}+3}{b+c}+\\frac{c^{2}+d^{2}+3}{c+d}+\\frac{d^{2}+a^{2}+3}{d+a} & \\geq 12-\\frac{a+b+c+d}{2} \\\\\n& \\geq 12-\\frac{4}{2}=10\n\\end{aligned}\n\)\nas required.\n\nEquality occurs if and only if $a = b = c = d = 1$, since then $a^2 + b^2 + c^2 + d^2 = 4$ and $a + b + c + d = 4$, so the minimum value of the sum is achieved and $C = 10$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 10$.

P300. Let $a, b, c, d, e$ be non-negative real numbers such that $a + b + c + d + e = 5$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c, d, e$ satisfying the given constraint: \(4\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\right)+5abcd \geq C\)

S300. $C = 25$

Without loss of generality we may assume that $a \geq b \geq c \geq d \geq e$.\nLet us denote\n\(\nf(a, b, c, d, e)=4\\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\\right)+5 a b c d\n\)\n\nThen we easily deduce that\n\(\n\\begin{equation*}\nf(a, b, c, d, e)-f\\left(\\frac{a+d}{2}, b, c, \\frac{a+d}{2}, e\\right)=\\frac{(a-d)^{2}}{4}(8-5 b c e) \\tag{1}\n\\end{equation*}\n\)\n\nSince $a \geq b \geq c \geq d \geq e$, we have\n\(\n3 \\sqrt[3]{b c e} \\leq b+c+e \\leq \\frac{3(a+b+c+d+e)}{5}=3\n\)\n\nThus it follows that bce $\leq 1$.\nNow, by (1) and the last inequality we get\n\(\n\\begin{aligned}\nf(a, b, c, d, e)-f\\left(\\frac{a+d}{2}, b, c, \\frac{a+d}{2}, e\\right) & =\\frac{(a-d)^{2}}{4}(8-5 b c e) \\\\\n& \\geq \\frac{(a-d)^{2}}{4}(8-5) \\geq 0\n\\end{aligned}\n\)\ni.e.\n\(\nf(a, b, c, d, e) \\geq f\\left(\\frac{a+d}{2}, b, c, \\frac{a+d}{2}, e\\right)\n\)\n\nAccording to the SMV theorem it remains to prove that $f(t, t, t, t, e) \geq 25$, under the condition $4 t+e=5$.\n\nClearly $4 t \leq 5$.\n\nWe have\n\(\n\\begin{aligned}\n& f(t, t, t, t, e) \\geq 25 \\\\\n& \\quad \\Leftrightarrow \\quad 4\\left(t^{2}+e^{2}\\right)+5 t^{4} e \\geq 25 \\\\\n& \\quad \\Leftrightarrow \\quad 4 t^{2}+4(5-4 t)^{2}+5 t^{4}(5-4 t)-25 \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad(5-4 t)(t-1)^{2}\\left(t^{2}+2 t+3\\right) \\geq 0\n\\end{aligned}\n\)\nwhich is true.\nEquality occurs if and only if $a=b=c=d=e=1$ or $a=b=c=d=5/4,\ e=0$ (up to permutation).\n\nEquality holds when either all variables are equal, i.e., $a = b = c = d = e = 1$, or when four variables are $5/4$ and the fifth is $0$, i.e., $a = b = c = d = 5/4$, $e = 0$ (up to permutation). In both cases, the expression attains its minimum value, so $C = 25$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 25$.

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