Inequality Proof Problems [301-350]
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Restructured the IneqMath training data.
P301. Let $a, b, c$ be positive real numbers. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: \(a^{b} b^{c} c^{a} \leq\left(C(a+b+c)\right)^{a+b+c}\)
S301. $C = \frac{1}{3}$
By the weighted power mean inequality we have\n\(\n\\begin{aligned}\n\\left(a^{b} b^{c} c^{a}\\right)^{\\frac{1}{a+b+c}} & =a^{\\frac{b}{a+b+c}} \\cdot b^{\\frac{c}{a+b+c}} \\cdot c^{\\frac{a}{a+b+c}} \\leq \\frac{b a+c b+a c}{a+b+c} \\leq \\frac{(a+b+c)^{2}}{3(a+b+c)} \\\\\n& =\\frac{a+b+c}{3}\n\\end{aligned}\n\)\n\nEquality holds when $a = b = c$, in which case $a^b b^c c^a = a^{a} a^{a} a^{a} = a^{3a}$ and $(C(a+b+c))^{a+b+c} = (3C a)^{3a}$, so $a^{3a} \leq (3C a)^{3a}$, which gives $C \geq \frac{1}{3}$. Thus, the minimal value of $C$ is $\frac{1}{3}$.\n\nTherefore, the answer is $C = \frac{1}{3}$.
P302. Let $a, b, c$ be the lengths of the sides of a triangle. Find the constant $C$ such that the following equation holds for all $a, b, c$: \(2(a b^{2} + b c^{2} + c a^{2}) = a^{2} b + b^{2} c + c^{2} a + C a b c\) and ensures that the triangle is equilateral.
S302. $C = 3$
We’ll show that\n\(\na^{2} b+b^{2} c+c^{2} a+3 a b c \\geq 2\\left(a b^{2}+b c^{2}+c a^{2}\\right)\n\)\nwith equality if and only if $a=b=c$, i.e. the triangle is equilateral.\nLet us use Ravi’s substitutions, i.e. $a=x+y, b=y+z, c=z+x$. Then the given inequality becomes\n\(\nx^{3}+y^{3}+z^{3}+x^{2} y+y^{2} z+z^{2} x \\geq 2\\left(x^{2} z+y^{2} x+z^{2} y\\right)\n\)\n\nSince $A M \geq G M$ we have\n\(\nx^{3}+z^{2} x \\geq 2 x^{2} z,\\quad y^{3}+x^{2} y \\geq 2 y^{2} x,\\quad z^{3}+y^{2} z \\geq 2 z^{2} y\n\)\n\nAfter adding these inequalities we obtain\n\(\nx^{3}+y^{3}+z^{3}+x^{2} y+y^{2} z+z^{2} x \\geq 2\\left(x^{2} z+y^{2} x+z^{2} y\\right)\n\)\n\nEquality holds if and only if $x = y = z$, which means $a = b = c$, i.e., the triangle is equilateral. This gives the unique value $C = 3$ for which the equation holds for all $a, b, c$ and ensures the triangle is equilateral.\n\nTherefore, the answer is $C = 3$.
P303. Let $D, E$ and $F$ be the feet of the altitudes of the triangle $ABC$ dropped from the vertices $A, B$ and $C$, respectively. Determine the largest constant $C$ such that the following inequality holds for all triangles $ABC$: \(\left(\frac{\overline{EF}}{a}\right)^{2}+\left(\frac{\overline{FD}}{b}\right)^{2}+\left(\frac{\overline{DE}}{c}\right)^{2} \geq C\)
S303. $C = \frac{3}{4}$
Clearly $\overline{E F}=a \cos \alpha, \overline{F D}=b \cos \beta, \overline{D E}=c \cos \gamma$, and the given inequality becomes\n\(\n\\cos ^{2} \\alpha+\\cos ^{2} \\beta+\\cos ^{2} \\gamma \\geq \\frac{3}{4}\n\)\n\nEquality holds when the triangle is equilateral, i.e., $\alpha = \beta = \gamma = 60^\circ$, so $\cos^2 60^\circ = (1/2)^2 = 1/4$ and the sum is $3 \times 1/4 = 3/4$. This gives the minimum value, so $C = \frac{3}{4}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{3}{4}$.
P304. Let $a, b, c > 0$ such that $abc = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \((a+b)(b+c)(c+a) \geq C(a+b+c-1)\)
S304. $C = 4$
We will use the fact that $(a+b)(b+c)(c+a) \geq \frac{8}{9}(a+b+c)(a b+b c+c a)$. So, it is enough to prove that $\frac{2}{9}(a b+b c+c a)+\frac{1}{a+b+c} \geq 1$. Using the AM-GM Inequality, we can write\n\n\(\n\\frac{2}{9}(a b+b c+c a)+\\frac{1}{a+b+c} \\geq 3 \\sqrt[3]{\\frac{(a b+b c+c a)^{2}}{81(a+b+c)}} \\geq 1\n\)\n\nbecause\n\n\(\n(a b+b c+c a)^{2} \\geq 3 a b c(a+b+c)=3(a+b+c)\n\)\n\nEquality holds when $a = b = c = 1$, since then $abc = 1$ and $(a+b)(b+c)(c+a) = 8$, $a+b+c-1 = 2$, so $C = 4$ is achieved. This gives the maximum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = 4$.”, “Using the identity $(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)-1$ we reduce the problem to the following one\n\n\(\na b+b c+c a+\\frac{3}{a+b+c} \\geq 4\n\)\n\nNow, we can apply the AM-GM Inequality in the following form\n\n\(\na b+b c+c a+\\frac{3}{a+b+c} \\geq 4 \\sqrt[4]{\\frac{(a b+b c+c a)^{3}}{9(a+b+c)}}\n\)\n\nAnd so it is enough to prove that\n\n\(\n(a b+b c+c a)^{3} \\geq 9(a+b+c)\n\)\n\nBut this is easy, because we clearly have $a b+b c+c a \geq 3$ and $(a b+b c+c a)^{2} \geq$ $3 a b c(a+b+c)=3(a+b+c)$.\n\nEquality holds when $a = b = c = 1$, since then $abc = 1$, $a+b+c = 3$, $ab+bc+ca = 3$, and $(a+b)(b+c)(c+a) = 8$, so $8 = 4(3-1)$. This gives the minimum value of $(a+b)(b+c)(c+a)$ for the given constraint, so $C = 4$ is the maximal constant.\n\nTherefore, the answer is $C = 4$.
P305. Let $a, b, c$ be positive real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: \(a b \frac{a+c}{b+c}+b c \frac{b+a}{c+a}+c a \frac{c+b}{a+b} \geq C \sqrt{a b c(a+b+c)}\)
S305. $C = \sqrt{3}$
Let $x=\frac{1}{b c}, y=\frac{1}{a c}, z=\frac{1}{a b}$ and $A=a c, B=a b, C=b c$.\nWe have\n\(\n\\begin{aligned}\n\\frac{x}{y+z}(B+C) & =a b \\frac{a+c}{b+c}, \\quad \\frac{y}{z+x}(C+A)=b c \\frac{b+a}{c+a} \\quad \\text { and } \\\\\n\\frac{z}{x+y}(A+B) & =c a \\frac{c+b}{a+b}\n\\end{aligned}\n\)\n\nUsing Corollary Let $a, b, c$ and $x, y, z$ be positive real numbers. Then\n\(\n\\frac{x}{y+z}(b+c)+\\frac{y}{z+x}(c+a)+\\frac{z}{x+y}(a+b) \\geq \\sqrt{3(a b+b c+c a)}\n\)\nand the previous identities we obtain\n\(\n\\begin{aligned}\n& a b \\frac{a+c}{b+c}+b c \\frac{b+a}{c+a}+c a \\frac{c+b}{a+b} \\\\\n& \\quad=\\frac{x}{y+z}(B+C)+\\frac{y}{z+x}(C+A)+\\frac{z}{x+y}(A+B) \\\\\n& \\quad \\geq \\sqrt{3(A B+B C+C A)}=\\sqrt{3 a b c(a+b+c)}\n\\end{aligned}\n\)\n\nEquality holds when $a = b = c$, since in this case all terms are equal and the inequality becomes an equality. This gives the minimum value of the left-hand side, so $C = \sqrt{3}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \sqrt{3}$.
P306. Let $a, b, x, y \in \mathbb{R}$ such that $a y - b x = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, x, y$ satisfying the given constraint: \(a^{2}+b^{2}+x^{2}+y^{2}+a x+b y \geq C.\)
S306. $C = \sqrt{3}$
Let us denote $u=a^{2}+b^{2}, v=x^{2}+y^{2}$ and $w=a x+b y$.\nThen\n\(\n\\begin{aligned}\nu v & =\\left(a^{2}+b^{2}\\right)\\left(x^{2}+y^{2}\\right)=a^{2} x^{2}+a^{2} y^{2}+b^{2} x^{2}+b^{2} y^{2} \\\\\n& =a^{2} x^{2}+b^{2} y^{2}+2 a x b y+a^{2} y^{2}+b^{2} x^{2}-2 a x b y \\\\\n& =(a x+b y)^{2}+(a y-b x)^{2}=w^{2}+1\n\\end{aligned}\n\)\n\nFrom the obvious inequality $(t \sqrt{3}+1)^{2} \geq 0$ we deduce\n\(\n3 t^{2}+1 \\geq-2 t \\sqrt{3}\n\)\ni.e.\n\(\n4 t^{2}+4 \\geq 3-2 t \\sqrt{3}+t^{2}\n\)\ni.e.\n\(\n\\begin{equation*}\n4 t^{2}+4 \\geq(\\sqrt{3}-t)^{2} \\tag{1}\n\\end{equation*}\n\)\n\nNow we have\n\(\n(u+v)^{2} \\geq 4 u v=4\\left(w^{2}+1\\right) \\stackrel{(1)}{\\geq}(\\sqrt{3}-w)^{2}\n\)\nfrom which we get $u+v \geq \sqrt{3}-w$, which is equivalent to $u+v+w \geq \sqrt{3}$.\n\nEquality holds when $a, b, x, y$ are such that $a y - b x = 1$ and $a x + b y = -\frac{\sqrt{3}}{2}$, with $a^2 + b^2 = x^2 + y^2 = \frac{\sqrt{3}}{2}$. This gives the minimum value of $a^2 + b^2 + x^2 + y^2 + a x + b y$, so $C = \sqrt{3}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \sqrt{3}$.
P307. Let $x, y, z$ be positive real numbers which satisfy the condition \(x y + x z + y z + 2 x y z = 1.\) Find the largest constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint: \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - C(x + y + z) \geq \frac{(2z-1)^2}{z(2z+1)}.\) , where $z=\max {x, y, z}$.
S307. $C = 4$
Of course, if $z$ is the greatest from the numbers $x, y, z$, then $z \geq \frac{1}{2}$; we saw that\n\n\(\n\\begin{aligned}\n\\frac{1}{x}+\\frac{1}{y}-4(x+y) & =(x+y)\\left(\\frac{1}{x y}-4\\right) \\geq \\frac{2}{2 z+1}(2 z+1) \\\\\n& =\\frac{2(2 z-1)(2 z+3)}{2 z+1}=4 z-\\frac{1}{z}+\\frac{(2 z-1)^{2}}{z(2 z+1)}\n\\end{aligned}\n\)\n\nfrom where we get the inequality\n\n\(\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}-4(x+y+z) \\geq \\frac{(2 z-1)^{2}}{z(2 z+1)}\n\)\n\nOf course, in the right-hand side $z$ could be replaced by any of the three numbers which is $\geq \frac{1}{2}$ (two such numbers could be, surely there is one).\n\nEquality holds when $x = y = z = \frac{1}{2}$, since then the constraint is satisfied and the inequality becomes an equality. This gives the maximum value of $C$ for which the inequality always holds, namely $C = 4$.\n\nTherefore, the answer is $C = 4$.
P308. Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: \(\sqrt{\frac{1}{a}-1} \sqrt{\frac{1}{b}-1} + \sqrt{\frac{1}{b}-1} \sqrt{\frac{1}{c}-1} + \sqrt{\frac{1}{c}-1} \sqrt{\frac{1}{a}-1} \geq C\)
S308. $C = 6$
Let $a=x y, b=y z, c=z x$. Then $x y+y z+z x=1$ and we may take\n\(\nx=\\tan \\frac{\\alpha}{2}, \\quad y=\\tan \\frac{\\beta}{2}, \\quad z=\\tan \\frac{\\gamma}{2}\n\)\nwhere $\alpha, \beta, \gamma \in(0, \pi)$ and $\alpha+\beta+\gamma=\pi$.\nWe have\n\(\n\\begin{aligned}\n\\sqrt{\\frac{1}{a}-1} \\sqrt{\\frac{1}{b}-1} & =\\sqrt{\\frac{(1-a)(1-b)}{a b}}=\\sqrt{\\frac{(1-x y)(1-y z)}{x y^{2} z}} \\\\\n& =\\sqrt{\\frac{(y z+z x)(z x+x y)}{x y^{2} z}}=\\sqrt{\\frac{(y+x)(z+y)}{y^{2}}}=\\frac{\\sqrt{1+y^{2}}}{y} \\\\\n& =\\frac{\\sqrt{1+\\tan ^{2} \\frac{\\beta}{2}}}{\\tan \\frac{\\beta}{2}}=\\frac{1}{\\sin \\frac{\\beta}{2}}\n\\end{aligned}\n\)\n\nSimilarly we obtain\n\(\n\\sqrt{\\frac{1}{b}-1} \\sqrt{\\frac{1}{c}-1}=\\frac{1}{\\sin \\frac{\\gamma}{2}} \\quad \\text { and } \\quad \\sqrt{\\frac{1}{c}-1} \\sqrt{\\frac{1}{a}-1}=\\frac{1}{\\sin \\frac{\\alpha}{2}}\n\)\n\nNow the given inequality becomes\n\(\n\\frac{1}{\\sin \\frac{\\alpha}{2}}+\\frac{1}{\\sin \\frac{\\beta}{2}}+\\frac{1}{\\sin \\frac{\\gamma}{2}} \\geq 6\n\)\n\nBy $A M \geq H M$ we have\n\(\n\\frac{1}{\\sin \\frac{\\alpha}{2}}+\\frac{1}{\\sin \\frac{\\beta}{2}}+\\frac{1}{\\sin \\frac{\\gamma}{2}} \\geq \\frac{9}{\\sin \\frac{\\alpha}{2}+\\sin \\frac{\\beta}{2}+\\sin \\frac{\\gamma}{2}}\n\)\n\nSo we need to prove that $\sin \frac{\alpha}{2}+\sin \frac{\beta}{2}+\sin \frac{\gamma}{2} \leq \frac{3}{2}$.\n\nEquality occurs if and only if $\alpha=\beta=\gamma=\pi / 3$, i.e. $a=b=c=\frac{1}{3}$. In this case, the minimum value of the expression is achieved, so $C = 6$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 6$.
P309. Let $x, y, z, t \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all positive $x, y, z, t$: \(x^{4}+y^{4}+z^{4}+t^{4}+C x y z t \geq x^{2} y^{2}+y^{2} z^{2}+z^{2} t^{2}+t^{2} x^{2}+x^{2} z^{2}+y^{2} t^{2}\)
S309. $C = 2$
Clearly, it is enough to prove the inequality if $x y z t=1$ and so the problem becomes\n\nIf $a, b, c, d$ have product 1 , then $a^{2}+b^{2}+c^{2}+d^{2}+2 \geq a b+b c+c d+d a+a c+b d$. Let $d$ the minimum among $a, b, c, d$ and let $m=\sqrt[3]{a b c}$. We will prove that\n$a^{2}+b^{2}+c^{2}+d^{2}+2-(a b+b c+c d+d a+a c+b d) \geq d^{2}+3 m^{2}+2-\left(3 m^{2}+3 m d\right)$, which is in fact\n\n\(\na^{2}+b^{2}+c^{2}-a b-b c-c a \\geq d(a+b+c-3 \\sqrt[3]{a b c})\n\)\n\nBecause $d \leq \sqrt[3]{a b c}$, proving this first inequality comes down to the inequality\n\n\(\na^{2}+b^{2}+c^{2}-a b-b c-c a \\geq \\sqrt[3]{a b c}(a+b+c-3 \\sqrt[3]{a b c})\n\)\n\nTake $u=\frac{a}{\sqrt[3]{a b c}}, v=\frac{b}{\sqrt[3]{a b c}}, w=\frac{c}{\sqrt[3]{a b c}}$. Using problem 74 , we find that\n\n\(\nu^{2}+v^{2}+w^{2}+3 \\geq u+v+w+u v+v w+w u\n\)\n\nwhich is exactly $a^{2}+b^{2}+c^{2}-a b-b c-c a \geq \sqrt[3]{a b c}(a+b+c-3 \sqrt[3]{a b c})$. Thus, it remains to prove that $d^{2}+2 \geq 3 m d \Leftrightarrow d^{2}+2 \geq 3 \sqrt[3]{d^{2}}$, which is clear.\n\nEquality holds when $x = y = z = t = 1$, since then both sides of the original inequality are equal: $4 + 2 \cdot 1 = 6$ and $6$. This gives the minimum value of $C$, so $C = 2$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.
P310. For any positive real numbers $x, y$ and any positive integers $m, n$, there exists a constant $C$ such that the following inequality holds: \(C\left(x^{m+n} + y^{m+n}\right) + (m+n-1)\left(x^m y^n + x^n y^m\right) \geq m n\left(x^{m+n-1} y + y^{m+n-1} x\right).\) Determine the optimal value of $C$.
S310. $C = (n-1)(m-1)$
We transform the inequality as follows:\n\n\(\n\\begin{gathered}\nm n(x-y)\\left(x^{m+n-1}-y^{m+n-1}\\right) \\geq(m+n-1)\\left(x^{m}-y^{m}\\right)\\left(x^{n}-y^{n}\\right) \\Leftrightarrow \\\\\n\\Leftrightarrow \\frac{x^{m+n-1}-y^{m+n-1}}{(m+n-1)(x-y)} \\geq \\frac{x^{m}-y^{m}}{m(x-y)} \\cdot \\frac{x^{n}-y^{n}}{n(x-y)}\n\\end{gathered}\n\)\n\n(we have assumed that $x>y$ ). The last relation can also be written\n\n\(\n(x-y) \\int_{y}^{x} t^{m+n-2} d t \\geq \\int_{y}^{x} t^{m-1} d t \\cdot \\int_{y}^{x} t^{n-1} d t\n\)\n\nand this follows from Chebyshev’s Inequality for integrals.\n\nEquality holds when $x = y$, since all terms become equal and the inequality becomes an equality. This gives the minimum value of $C$, so the optimal value is $C = (n-1)(m-1)$.\n\nTherefore, the answer is $C = (n-1)(m-1)$.
P311. Let $a, b, c$ be positive real numbers 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: \(\left(a^{a}+b^{a}+c^{a}\right)\left(a^{b}+b^{b}+c^{b}\right)\left(a^{c}+b^{c}+c^{c}\right) \geq C(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c})^{3}\)
S311. $C = 1$
By Hölder’s inequality we obtain\n\(\n\\left(a^{a}+b^{a}+c^{a}\\right)^{\\frac{1}{3}}\\left(a^{b}+b^{b}+c^{b}\\right)^{\\frac{1}{3}}\\left(a^{c}+b^{c}+c^{c}\\right)^{\\frac{1}{3}} \\geq a^{\\frac{a+b+c}{3}}+b^{\\frac{a+b+c}{3}}+c^{\\frac{a+b+c}{3}}\n\)\n\nSince $a+b+c=1$, the conclusion follows.\n\nEquality holds when $a = b = c = \frac{1}{3}$, in which case both sides of the inequality are equal. 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 $C = 1$.
P312. Let $a, b, c \in \mathbb{R}^{+}$ such that $abc = 1$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$: \(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leq C \left( \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c} \right)\)
S312. $C = 1$
Let $x=a+b+c$ and $y=a b+b c+c a$. Using brute-force, it is easy to see that the left hand side is $\frac{x^{2}+4 x+y+3}{x^{2}+2 x+y+x y}$, while the right hand side is $\frac{12+4 x+y}{9+4 x+2 y}$. Now, the inequality becomes\n\n\(\n\\frac{x^{2}+4 x+y+3}{x^{2}+2 x+y+x y}-1 \\leq \\frac{12+4 x+y}{9+4 x+2 y}-1 \\Leftrightarrow \\frac{2 x+3-x y}{x^{2}+2 x+y+x y} \\leq \\frac{3-y}{9+4 x+2 y}\n\)\n\nFor the last inequality, we clear denominators. Then using the inequalities $x \geq 3, y \geq 3, x^{2} \geq 3 y$, we have\n\n\(\n\\frac{5}{3} x^{2} y \\geq 5 x^{2}, \\frac{x^{2} y}{3} \\geq y^{2}, x y^{2} \\geq 9 x, 5 x y \\geq 15 x, x y \\geq 3 y \\text { and } x^{2} y \\geq 27\n\)\n\nSumming up these inequalities, the desired inequality follows.\n\nEquality holds when $a = b = c = 1$, since then $abc = 1$ and both sides of the original inequality are equal. 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 $C = 1$.
P313. 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 $a, b, c$: \(\left(1 + a + a^2\right)\left(1 + b + b^2\right)\left(1 + c + c^2\right) \geq C(ab + bc + ca).\)
S313. $C = 9$
Let us denote $x=a+b+c=3, y=a b+b c+c a, z=a b c$.\nNow the given inequality can be rewritten as\n\(\nz^{2}-2 z-2 x z+z(x+y)+x^{2}+x+y^{2}-y+3 x y+1 \\geq 9 y\n\)\ni.e.\n\(\n(z-1)^{2}-(z-1)(x-y)+(x-y)^{2} \\geq 0\n\)\nwhich is obviously true. Equality holds iff $a=b=c=1$.\n\nEquality is achieved when $a = b = c = 1$, in which case $ab + bc + ca = 3$ and $\left(1 + a + a^2\right)^3 = 27$, so the inequality becomes $27 \geq 9 \times 3 = 27$. Thus, $C = 9$ is the largest constant for which the inequality always holds, and this gives the minimum value of $C$.\n\nTherefore, the answer is $C = 9$.
P314. Let $a, b, c \in (-1, 1)$ be real numbers such that $a b + b c + a c = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(C \sqrt[3]{\left(1-a^{2}\right)\left(1-b^{2}\right)\left(1-c^{2}\right)} \leq 1 + (a + b + c)^2\)
S314. $C = 6$
Since $a, b, c \in(-1,1)$ we have $1-a^{2}, 1-b^{2}, 1-c^{2}>0$.\nBy $A M \geq G M$ we get\n\(\n\\begin{aligned}\n6 \\sqrt[3]{\\left(1-a^{2}\\right)\\left(1-b^{2}\\right)\\left(1-c^{2}\\right)} & =2 \\cdot 3 \\sqrt[3]{\\left(1-a^{2}\\right)\\left(1-b^{2}\\right)\\left(1-c^{2}\\right)} \\\\\n& \\leq 2\\left(1-a^{2}+1-b^{2}+1-c^{2}\\right) \\\\\n& =2\\left(3-\\left(a^{2}+b^{2}+c^{2}\\right)\\right) \\\\\n& =6-2\\left(a^{2}+b^{2}+c^{2}\\right)\n\\end{aligned}\n\)\n\nWe’ll show that\n\(\n6-2\\left(a^{2}+b^{2}+c^{2}\\right) \\leq 1+(a+b+c)^{2}\n\)\n\nThis inequality is equivalent to\n\(\n6-2\\left(a^{2}+b^{2}+c^{2}\\right) \\leq 1+a^{2}+b^{2}+c^{2}+2\n\)\ni.e.\n\(\n3 \\leq 3\\left(a^{2}+b^{2}+c^{2}\\right)\n\)\ni.e.\n\(\na^{2}+b^{2}+c^{2} \\geq 1\n\)\nwhich is true since $a^{2}+b^{2}+c^{2} \geq a b+b c+a c=1$.\n\nEquality holds when $a = b = c = \pm \frac{1}{\sqrt{3}}$, since then $ab + bc + ca = 1$ and $a^2 + b^2 + c^2 = 1$. In this case, the inequality becomes equality, so $C = 6$ is the largest possible value.\n\nTherefore, the answer is $C = 6$.
P315. Let $x, y, z > 0$ satisfy the condition $x + y + z = xyz$. Find the maximal constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given condition: \(xy + xz + yz \geq C + \sqrt{x^2 + 1} + \sqrt{y^2 + 1} + \sqrt{z^2 + 1}\)
S315. $C = 3$
Another improvement is as follows. Start from\n\n\(\n\\frac{1}{x^{2}}+\\frac{1}{y^{2}}+\\frac{1}{z^{2}} \\geq \\frac{1}{x y}+\\frac{1}{x z}+\\frac{1}{y z}=1 \\Rightarrow x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2} \\geq x^{2} y^{2} z^{2}\n\)\n\nwhich is equivalent to\n\n\(\n(x y+x z+y z)^{2} \\geq 2 x y z(x+y+z)+x^{2} y^{2} z^{2}=3(x+y+z)^{2}\n\)\n\nFurther on,\n\n\(\n\\begin{aligned}\n& (x y+x z+y z-3)^{2}=(x y+x z+y z)^{2}-6(x y+x z+y z)+9 \\geq \\\\\n& \\geq 3(x+y+z)^{2}-6(x y+x z+y z)+9=3\\left(x^{2}+y^{2}+z^{2}\\right)+9\n\\end{aligned}\n\)\n\nso that\n\n\(\nx y+x z+y z \\geq 3+\\sqrt{3\\left(x^{2}+y^{2}+z^{2}\\right)+9}\n\)\n\nBut\n\n\(\n\\sqrt{3\\left(x^{2}+y^{2}+z^{2}\\right)+9} \\geq \\sqrt{x^{2}+1}+\\sqrt{y^{2}+1}+\\sqrt{z^{2}+1}\n\)\n\nis a consequence of the Cauchy-Schwarz Inequality and we have a second improvement and proof for the desired inequality:\n\n\(\n\\begin{aligned}\nx y+x z+y z & \\geq 3+\\sqrt{3\\left(x^{2}+y^{2}+z^{2}\\right)+9} \\geq \\\\\n& \\geq 3+\\sqrt{x^{2}+1}+\\sqrt{y^{2}+1}+\\sqrt{z^{2}+1}\n\\end{aligned}\n\)\n\nEquality holds when $x = y = z = \sqrt{3}$, since then $x + y + z = 3\sqrt{3}$ and $xyz = (\sqrt{3})^3 = 3\sqrt{3}$, so the condition is satisfied. In this case, $xy + xz + yz = 3 \cdot 3 = 9$ and $\sqrt{x^2+1} = \sqrt{3+1} = 2$ for each variable, so $C = 9 - 3 \times 2 = 3$. This gives the minimum value of $xy + xz + yz - (\sqrt{x^2+1} + \sqrt{y^2+1} + \sqrt{z^2+1})$, so the maximal constant $C$ is $3$.\n\nTherefore, the answer is $C = 3$. “, “We have\n\n\(\nx y z=x+y+z \\geq 2 \\sqrt{x y}+z \\Rightarrow z(\\sqrt{x y})^{2}-2 \\sqrt{x y}-z \\geq 0\n\)\n\nBecause the positive root of the trinomial $z t^{2}-2 t-z$ is\n\n\(\n\\frac{1+\\sqrt{1+z^{2}}}{z}\n\)\n\nwe get from here\n\n\(\n\\sqrt{x y} \\geq \\frac{1+\\sqrt{1+z^{2}}}{z} \\Leftrightarrow z \\sqrt{x y} \\geq 1+\\sqrt{1+z^{2}}\n\)\n\nOf course, we have two other similar inequalities. Then,\n\n\(\n\\begin{aligned}\nx y+x z+y z & \\geq x \\sqrt{y z}+y \\sqrt{x z}+z \\sqrt{x y} \\geq \\\\\n& \\geq 3+\\sqrt{x^{2}+1}+\\sqrt{y^{2}+1}+\\sqrt{z^{2}+1}\n\\end{aligned}\n\)\n\nand we have both a proof of the given inequality, and a little improvement of it.\n\nEquality holds when $x = y = z = 1$, since then $x + y + z = 3 = xyz$, and $xy + yz + zx = 3$, $\sqrt{x^2+1} + \sqrt{y^2+1} + \sqrt{z^2+1} = 3\sqrt{2}$, so the inequality becomes $3 \geq 3 + 3\sqrt{2}$, which is tight for $C = 3$. This gives the minimum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = 3$.
P316. Let $a, b, c \in \mathbb{R}^{+}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: \(\frac{a}{b+2 c}+\frac{b}{c+2 a}+\frac{c}{a+2 b} \geq C\)
S316. $C = 1$
Applying the Cauchy-Schwarz inequality we get\n\(\n\\begin{aligned}\n& \\left(\\frac{a}{b+2 c}+\\frac{b}{c+2 a}+\\frac{c}{a+2 b}\\right)(a(b+2 c)+b(c+2 a)+c(a+2 b)) \\\\\n& \\quad \\geq(a+b+c)^{2}\n\\end{aligned}\n\)\nhence\n\(\n\\frac{a}{b+2 c}+\\frac{b}{c+2 a}+\\frac{c}{a+2 b} \\geq \\frac{(a+b+c)^{2}}{3(a b+b c+c a)}\n\)\n\nSo it suffices to show that\n\(\n\\frac{(a+b+c)^{2}}{3(a b+b c+c a)} \\geq 1, \\quad \\text { i.e. } \\quad(a+b+c)^{2} \\geq 3(a b+b c+c a)\n\)\nwhich is equivalent to $a^{2}+b^{2}+c^{2} \geq a b+b c+c a$, and clearly holds.\nEquality occurs iff $a=b=c$.\n\nEquality holds when $a = b = c$, in which case $\frac{a}{b+2c} + \frac{b}{c+2a} + \frac{c}{a+2b} = 1$. 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 $C = 1$.
P317. Let $a, b, c$ be the lengths of the sides of a triangle, such that $a + b + c = 3$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$: \(a^{2} + b^{2} + c^{2} + \frac{4 a b c}{3} \geq C.\)
S317. $C = \frac{13}{3}$
Let $a=x+y, b=y+z$ and $c=z+x$.\nSo we have $x+y+z=\frac{3}{2}$ and since $A M \geq G M$ we get $x y z \leq\left(\frac{x+y+z}{3}\right)^{3}=\frac{1}{8}$.\n\nNow we obtain\n\(\n\\begin{aligned}\na^{2} & +b^{2}+c^{2}+\\frac{4 a b c}{3} \\\\\n& =\\frac{\\left(a^{2}+b^{2}+c^{2}\\right)(a+b+c)+4 a b c}{3} \\\\\n& =\\frac{2\\left((x+y)^{2}+(y+z)^{2}+(z+x)^{2}\\right)(x+y+z)+4(x+y)(y+z)(z+x)}{3} \\\\\n& =\\frac{4}{3}\\left((x+y+z)^{3}-x y z\\right) \\geq \\frac{4}{3}\\left(\\left(\\frac{3}{2}\\right)^{3}-\\frac{1}{8}\\right)=\\frac{13}{3}\n\\end{aligned}\n\)\n\nEquality holds when $x = y = z$, which means $a = b = c = 1$. In this case, the minimum value of the expression is achieved, so $C = \frac{13}{3}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{13}{3}$.
P318. Let $a, b, c \in \mathbb{R}^{+}$ with $a^{2}+b^{2}+c^{2}=3$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$: \(\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b} \geq C.\)
S318. $C = 3$
The given inequality is equivalent to\n\(\n\\begin{aligned}\n& \\left(\\frac{a b}{c}+\\frac{b c}{a}+\\frac{c a}{b}\\right)^{2} \\geq 9 \\\\\n& \\quad \\Leftrightarrow \\quad \\frac{a^{2} b^{2}}{c^{2}}+\\frac{b^{2} c^{2}}{a^{2}}+\\frac{c^{2} a^{2}}{b^{2}}+2\\left(a^{2}+b^{2}+c^{2}\\right) \\geq 3\\left(a^{2}+b^{2}+c^{2}\\right)\n\\end{aligned}\n\)\ni.e.\n\(\n\\frac{a^{2} b^{2}}{c^{2}}+\\frac{b^{2} c^{2}}{a^{2}}+\\frac{c^{2} a^{2}}{b^{2}} \\geq a^{2}+b^{2}+c^{2}\n\)\n\nFurthermore, applying $A M \geq G M$ we get\n\(\n\\frac{a^{2} b^{2}}{c^{2}}+\\frac{b^{2} c^{2}}{a^{2}} \\geq 2 b^{2}, \\quad \\frac{b^{2} c^{2}}{a^{2}}+\\frac{c^{2} a^{2}}{b^{2}} \\geq 2 c^{2}, \\quad \\frac{a^{2} b^{2}}{c^{2}}+\\frac{c^{2} a^{2}}{b^{2}} \\geq 2 a^{2}\n\)\n\nAfter adding these inequalities we obtain\n\(\n\\frac{a^{2} b^{2}}{c^{2}}+\\frac{b^{2} c^{2}}{a^{2}}+\\frac{c^{2} a^{2}}{b^{2}} \\geq a^{2}+b^{2}+c^{2}\n\)\nand we are done.\n\nEquality holds when $a = b = c = 1$, since then $a^2 + b^2 + c^2 = 3$ and $\frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b} = 3$. This gives the minimum value of the expression, so the largest constant $C$ for which the inequality always holds is $C = 3$.\n\nTherefore, the answer is $C = 3$.
P319. Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$: \(\frac{a}{a + bc} + \frac{b}{b + ca} + \frac{\sqrt{abc}}{c + ab} \leq C.\)
S319. $C = 1 + \frac{3\sqrt{3}}{4}$
Since $a+b+c=1$ we use the following substitutions $a=xy, b=yz, c=zx$, where $x, y, z>0$ and the given inequality becomes\n\(\n\\frac{xy}{xy+(yz)(zx)}+\\frac{yz}{yz+(zx)(xy)}+\\frac{xyz}{zx+(xy)(yz)} \\leq 1+\\frac{3 \\sqrt{3}}{4}\n\)\ni.e.\n\(\n\\begin{equation*}\n\\frac{1}{1+z^{2}}+\\frac{1}{1+x^{2}}+\\frac{y}{1+y^{2}} \\leq 1+\\frac{3 \\sqrt{3}}{4} \\tag{1}\n\\end{equation*}\n\)\nwhere $xy+yz+zx=1$.\n\nSince $xy+yz+zx=1$ according to Case 3 we may set $x=\tan \frac{\alpha}{2}, y=\tan \frac{\beta}{2}, z=\tan \frac{\gamma}{2}$ where $\alpha, \beta, \gamma \in(0, \pi)$, and $\alpha+\beta+\gamma=\pi$.\n\nThen inequality (1) becomes\n\(\n\\frac{1}{1+\\tan ^{2} \\frac{\\gamma}{2}}+\\frac{1}{1+\\tan ^{2} \\frac{\\alpha}{2}}+\\frac{\\tan \\frac{\\beta}{2}}{1+\\tan ^{2} \\frac{\\beta}{2}} \\leq 1+\\frac{3 \\sqrt{3}}{4}\n\)\ni.e.\n\(\n\\cos ^{2} \\frac{\\gamma}{2}+\\cos ^{2} \\frac{\\alpha}{2}+\\frac{\\sin \\beta}{2} \\leq 1+\\frac{3 \\sqrt{3}}{4}\n\)\n\nUsing the trigonometric identity $\cos x=2 \cos ^{2} \frac{x}{2}-1$ the last inequality becomes\n\(\n\\frac{\\cos \\gamma+1}{2}+\\frac{\\cos \\alpha+1}{2}+\\frac{\\sin \\beta}{2} \\leq 1+\\frac{3 \\sqrt{3}}{4}\n\)\ni.e.\n\(\n\\begin{equation*}\n\\cos \\gamma+\\cos \\alpha+\\sin \\beta \\leq \\frac{3 \\sqrt{3}}{2} \\tag{2}\n\\end{equation*}\n\)\n\nWe have\n\(\n\\begin{aligned}\n\\cos \\alpha+\\cos \\gamma+\\sin \\beta= & \\cos \\alpha+\\cos \\gamma+\\sin (\\pi-(\\alpha+\\gamma)) \\\\\n= & \\frac{2}{\\sqrt{3}}\\left(\\frac{\\sqrt{3}}{2} \\cos \\alpha+\\frac{\\sqrt{3}}{2} \\cos \\gamma\\right) \\\\\n& +\\frac{1}{\\sqrt{3}}(\\sqrt{3} \\sin \\alpha \\cos \\gamma+\\sqrt{3} \\cos \\alpha \\sin \\gamma) \\\\\n\\leq & \\frac{1}{\\sqrt{3}}\\left(\\frac{3}{4}+\\cos ^{2} \\alpha+\\frac{3}{4}+\\cos ^{2} \\gamma\\right) \\\\\n= & \\frac{1}{2 \\sqrt{3}}\\left(3 \\sin ^{2} \\alpha+\\cos ^{2} \\gamma+\\cos ^{2} \\alpha+3 \\sin ^{2} \\gamma\\right) \\\\\n= & \\frac{\\sqrt{3}}{2}+\\frac{\\sqrt{3}}{2}\\left(\\cos ^{2} \\alpha+\\sin ^{2} \\alpha\\right)+\\frac{\\sqrt{3}}{2}\\left(\\cos ^{2} \\gamma+\\sin ^{2} \\gamma\\right) \\\\\n= & \\frac{3 \\sqrt{3}}{2} .\n\\end{aligned}\n\)\n\nEquality in the above inequalities is achieved when $\alpha = \gamma = \frac{\pi}{3}$ and $\beta = \frac{\pi}{3}$, which corresponds to $a = b = c = \frac{1}{3}$. In this case, the left-hand side attains its maximum value, so $C = 1 + \frac{3\sqrt{3}}{4}$ is the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1 + \frac{3\sqrt{3}}{4}$.
P320. Let $x, y, z$ be positive real numbers. Find the minimal constant $C$ such that the following inequality holds for all $x, y, z \in \mathbb{R}^{+}$: \(\frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}} \leq C\)
S320. $C = 1$
From Huygens Inequality we have $\sqrt{(x+y)(x+z)} \geq x+\sqrt{y z}$ and using this inequality for the similar ones we get\n\n\(\n\\begin{aligned}\n\\frac{x}{x+\\sqrt{(x+y)(x+z)}} & +\\frac{y}{y+\\sqrt{(y+z)(y+x)}}+\\frac{z}{z+\\sqrt{(z+x)(z+y)}} \\leq \\\\\n& \\leq \\frac{x}{2 x+\\sqrt{y z}}+\\frac{y}{2 y+\\sqrt{z x}}+\\frac{z}{2 z+\\sqrt{x y}}\n\\end{aligned}\n\)\n\nNow, we denote with $a=\frac{\sqrt{y z}}{x}, b=\frac{\sqrt{z x}}{y}, c=\frac{\sqrt{x y}}{z}$ and the inequality becomes\n\n\(\n\\frac{1}{2+a}+\\frac{1}{2+b}+\\frac{1}{2+c} \\leq 1\n\)\n\nFrom the above notations we can see that $a b c=1$, so the last inequality becomes after clearing the denominators $a b+b c+c a \geq 3$, which follows from the AM-GM Inequality.\n\nEquality holds when $x = y = z$, in which case $a = b = c = 1$ and $\frac{1}{2+1} + \frac{1}{2+1} + \frac{1}{2+1} = 1$. This gives the maximum value of the original expression, so $C = 1$ is the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$. “, “We have $(x+y)(x+z)=x y+\left(x^{2}+y z\right)+x z \geq x y+2 x \sqrt{y z}+x z=(\sqrt{x y}+\sqrt{x z})^{2}$. Hence\n\n\(\n\\sum \\frac{x}{x+\\sqrt{(x+y)(x+z)}} \\leq \\sum \\frac{x}{x+\\sqrt{x y}+\\sqrt{x z}}\n\)\n\nBut\n\n\(\n\\sum \\frac{x}{x+\\sqrt{x y}+\\sqrt{x z}}=\\sum \\frac{\\sqrt{x}}{\\sqrt{x}+\\sqrt{y}+\\sqrt{z}}=1\n\)\n\nand this solves the problem.\n\nEquality holds when $x = y = z$, since then each term becomes $\frac{x}{x + x + x} = \frac{1}{3}$, and the sum is $1$. This gives the maximum value of the sum, so $C = 1$ is minimal such that the inequality always holds.\n\nTherefore, the answer is $C = 1$.
P321. Let $a, b, c$ be non-negative real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: \(\frac{a}{4 b^{2}+b c+4 c^{2}}+\frac{b}{4 c^{2}+c a+4 a^{2}}+\frac{c}{4 a^{2}+a b+4 b^{2}} \geq C \cdot \frac{1}{a+b+c}\)
S321. $C = 1$
By the Cauchy-Schwarz inequality we have\n\(\n\\begin{gathered}\n\\frac{a}{4 b^{2}+b c+4 c^{2}}+\\frac{b}{4 c^{2}+c a+4 a^{2}}+\\frac{c}{4 a^{2}+a b+4 b^{2}} \\\\\n\\geq \\frac{(a+b+c)^{2}}{4 a\\left(b^{2}+c^{2}\\right)+4 b\\left(c^{2}+a^{2}\\right)+4 c\\left(a^{2}+b^{2}\\right)+3 a b c}\n\\end{gathered}\n\)\n\nSo we need to prove that\n\(\n\\frac{(a+b+c)^{2}}{4 a\\left(b^{2}+c^{2}\\right)+4 b\\left(c^{2}+a^{2}\\right)+4 c\\left(a^{2}+b^{2}\\right)+3 a b c} \\geq \\frac{1}{a+b+c}\n\)\nwhich is equivalent to\n\(\n(a+b+c)^{3} \\geq 4 a\\left(b^{2}+c^{2}\\right)+4 b\\left(c^{2}+a^{2}\\right)+4 c\\left(a^{2}+b^{2}\\right)+3 a b c\n\)\ni.e.\n\(\na^{3}+b^{3}+c^{3}+3 a b c \\geq a\\left(b^{2}+c^{2}\\right)+b\\left(c^{2}+a^{2}\\right)+c\\left(a^{2}+b^{2}\\right)\n\)\nwhich is Schur’s inequality.\n\nEquality holds when $a = b = c$, for example at $a = b = c > 0$. In this case, both sides of the inequality are equal, so the minimum value of the left side is achieved and $C = 1$ is the largest possible constant.\n\nTherefore, the answer is $C = 1$.
P322. Let $a, b, c \in \mathbb{R}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: \(a^{2}+b^{2}+c^{2} \geq C(a b+b c+c a)\)
S322. $C = 1$
Since $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$ we deduce\n\(\n2\\left(a^{2}+b^{2}+c^{2}\\right) \\geq 2(a b+b c+c a) \\Leftrightarrow a^{2}+b^{2}+c^{2} \\geq a b+b c+c a\n\)\n\nEquality occurs if and only if $a = b = c$, in which case both sides are equal. This gives the minimum value of $a^2 + b^2 + c^2 - C(ab + bc + ca)$, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.
P323. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots+a_{n}=1$. Find the largest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$: \(\frac{a_{1}}{\sqrt{1-a_{1}}}+\frac{a_{2}}{\sqrt{1-a_{2}}}+\cdots+\frac{a_{n}}{\sqrt{1-a_{n}}} \geq C\)
S323. $C = \sqrt{\frac{n}{n-1}}$
Let us denote\n\(\n\\begin{aligned}\n& A=\\frac{a_{1}}{\\sqrt{1-a_{1}}}+\\frac{a_{2}}{\\sqrt{1-a_{2}}}+\\cdots+\\frac{a_{n}}{\\sqrt{1-a_{n}}} \\\\\n& B=a_{1}\\left(1-a_{1}\\right)+a_{2}\\left(1-a_{2}\\right)+\\cdots+a_{n}\\left(1-a_{n}\\right)\n\\end{aligned}\n\)\n\nBy Hölder’s inequality we have\n\(\n\\begin{equation*}\nA^{2} B \\geq\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{3}=1 \\tag{1}\n\\end{equation*}\n\)\n\nApplying $QM \geq AM$ we deduce\n\(\n\\begin{equation*}\nB=1-\\left(a_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2}\\right) \\leq 1-\\frac{\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2}}{n}=\\frac{n-1}{n} \\tag{2}\n\\end{equation*}\n\)\n\nBy (1) and (2) we obtain\n\(\n\\frac{n-1}{n} \\cdot A^{2} \\geq A^{2} B \\geq 1, \\quad \\text { i.e. } \\quad A \\geq \\sqrt{\\frac{n}{n-1}}\n\)\n\nEquality holds when $a_{1} = a_{2} = \cdots = a_{n} = \frac{1}{n}$, since then both Hölder’s and QM-AM inequalities become equalities. This gives the minimum value of $A$, so $C = \sqrt{\frac{n}{n-1}}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \sqrt{\frac{n}{n-1}}$.
P324. Let $a, b, c$ be real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}$: \(\left(a^{2}+b^{2}+c^{2}\right)^{2} \geq C\left(a^{3} b+b^{3} c+c^{3} a\right)\)
S324. $C = 3$
By the well-known inequality $(x+y+z)^{2} \geq 3(xy+yz+zx)$ for\n\(\nx = a^{2} + b c - a b, \\quad y = b^{2} + c a - b c, \\quad z = c^{2} + a b - c a\n\)\nwe obtain the required inequality.\n\nEquality holds when $a = b = c$, in which case both sides of the inequality are equal. This gives the maximum value of $C$ for which the inequality always holds, namely $C = 3$.\n\nTherefore, the answer is $C = 3$.
P325. 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 $a, b, c$ satisfying the given constraint: \((a+b)^{2}(1+2 c)(2 a+3 c)(2 b+3 c) \geq C a b c\)
S325. $C = 54$
The given inequality can be rewritten as follows\n\(\n(a+b)^{2}(1+2 c)\\left(2+3 \\frac{c}{a}\\right)\\left(2+3 \\frac{c}{b}\\right) \\geq 54 c\n\)\n\nBy the Cauchy-Schwarz inequality and $A M \geq G M$ we have\n\(\n\\begin{aligned}\n\\left(2+3 \\frac{c}{a}\\right)\\left(2+3 \\frac{c}{b}\\right) & \\geq\\left(2+\\frac{3 c}{\\sqrt{a b}}\\right)^{2} \\geq\\left(2+\\frac{6 c}{a+b}\\right)^{2}=\\frac{(2(a+b)+6 c)^{2}}{(a+b)^{2}} \\\\\n& =\\frac{4(1+2 c)^{2}}{(a+b)^{2}}\n\\end{aligned}\n\)\n\nThen we have\n\(\n\\begin{aligned}\n(a+b)^{2}(1+2 c)\\left(2+3 \\frac{c}{a}\\right)\\left(2+3 \\frac{c}{b}\\right) & \\geq(a+b)^{2}(1+2 c) \\frac{4(1+2 c)^{2}}{(a+b)^{2}} \\\\\n& =4(1+2 c)^{3}\n\\end{aligned}\n\)\nand it remains to prove that\n\(\n4(1+2 c)^{3} \\geq 54 c, \\quad \\text { i.e. } \\quad(1+2 c)^{3} \\geq \\frac{27 c}{2}\n\)\n\nBy the $A M \geq G M$ inequality we have\n\(\n(1+2 c)^{3}=\\left(\\frac{1}{2}+\\frac{1}{2}+2 c\\right)^{3} \\geq 27 \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot 2 c=\\frac{27 c}{2}\n\)\nas required.\nEquality occurs iff $a=b=\frac{3}{8},\ c=\frac{1}{4}$, that is, when $a$ and $b$ are equal and $c$ is $\frac{1}{4}$. This gives the maximum value of $C$ for which the inequality always holds, so $C=54$ is maximal.\n\nTherefore, the answer is $C = 54$.
P326. Let $a, b, c > 0$ be real numbers. Determine the maximal constant $C$ such that the following inequality holds for all $a, b, c$: \((a b + b c + c a) \left( \frac{1}{(a + b)^2} + \frac{1}{(b + c)^2} + \frac{1}{(c + a)^2} \right) \geq C.\)
S326. $C = \frac{9}{4}$
We can rewrite the given inequality in the following form\n\(\n\\begin{aligned}\n& f(a+b+c, a b+b c+c a, a b c) \\\\\n&= 9((a+b)(b+c)(c+a))^{2} \\\\\n&-4(a b+b c+c a)\\left((a+b)^{2}(b+c)^{2}+(b+c)^{2}(c+a)^{2}+(c+a)^{2}(a+b)^{2}\\right) \\\\\n&= k(a b c)^{2}+m a b c+n\n\\end{aligned}\n\)\nwhere $k \geq 0$ and $k, m, n$ are quantities containing constants or $a+b+c, a b+$ $b c+c a, a b c$, which we also consider as constants, i.e. in the form as a sixth-degree symmetric polynomial with variables $a, b, c$ and a second-degree polynomial with variable $a b c$ and positive coefficients.\n\nLet us explain this:\nThe expression $(a+b)(b+c)(c+a)$ has the form $k a b c+m$ so it follows that $9((a+b)(b+c)(c+a))^{2}$ has the form $k^{2}(a b c)^{2}+m a b c+n$.\n\nFurthermore\n$4(a b+b c+c a)\left((a+b)^{2}(b+c)^{2}+(b+c)^{2}(c+a)^{2}+(c+a)^{2}(a+b)^{2}\right)=4 k A$,\nwhere $k=a b+b c+c a$, and $A$ is a fourth-degree polynomial and also has the form $k a b c+m$.\n\nTherefore the expression of the left side of $f(a+b+c, a b+b c+c a, a b c)$ has the form $k(a b c)^{2}+m a b c+n$.\n\nThen the function achieves it’s minimum value when $(a-b)(b-c)(c-a)=0$ or when $a b c=0$.\n\nIf $(a-b)(b-c)(c-a)=0$, then without loss of generality we may assume that $a=c$, and the given inequality is equivalent to\n\(\n\\begin{aligned}\n& \\left(a^{2}+2 a b\\right)\\left(\\frac{1}{4 a^{2}}+\\frac{2}{(a+b)^{2}}\\right) \\geq \\frac{9}{4} \\\\\n& \\quad \\Leftrightarrow \\quad(a-b)^{2}\\left(\\frac{2 a+b}{2 a(a+b)^{2}}-\\frac{1}{(a+b)^{2}}\\right) \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad b(a-b)^{2} \\geq 0\n\\end{aligned}\n\)\nas required.\nIf $a b c=0$, we may assume that $c=0$ and the given inequality becomes\n\(\n\\begin{aligned}\n& a b\\left(\\frac{1}{(a+b)^{2}}+\\frac{1}{a^{2}}+\\frac{1}{b^{2}}\\right) \\geq \\frac{9}{4} \\quad \\Leftrightarrow \\quad(a-b)^{2}\\left(\\frac{1}{a b}-\\frac{1}{4(a+b)^{2}}\\right) \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad(a-b)^{2}\\left(4 a^{2}+4 b^{2}+7 a b\\right) \\geq 0\n\\end{aligned}\n\)\nand the problem is solved.\n\nEquality in the above analysis holds when two variables are equal and the third is zero, i.e., when $(a, b, c)$ is a permutation of $(t, t, 0)$ for $t > 0$. In this case, the minimum value of the expression is achieved, so $C = \frac{9}{4}$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{9}{4}$.
P327. Let $k \in \mathbb{N}$, and $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1} + a_{2} + \cdots + a_{n} = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$: \(a_{1}^{-k} + a_{2}^{-k} + \cdots + a_{n}^{-k} \geq C.\)
S327. $C = n^{k+1}$
Since $A M \geq G M$ we have\n\(\n\\sqrt[n]{a_{1} a_{2} \\cdots a_{n}} \\leq \\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}=\\frac{1}{n}\n\)\nor\n\(\nn \\leq \\sqrt[n]{\\frac{1}{a_{1}} \\frac{1}{a_{2}} \\cdots \\frac{1}{a_{n}}}\n\)\n\nHence\n\(\nn^{k} \\leq \\sqrt[n]{a_{1}^{-k} a_{2}^{-k} \\cdots a_{n}^{-k}} \\leq \\frac{a_{1}^{-k}+a_{2}^{-k}+\\cdots+a_{n}^{-k}}{n}\n\)\ni.e.\n\(\na_{1}^{-k}+a_{2}^{-k}+\\cdots+a_{n}^{-k} \\geq n^{k+1}\n\)\nas required.\n\nEquality holds when $a_1 = a_2 = \cdots = a_n = \frac{1}{n}$, since then $a_i^{-k} = n^k$ for each $i$, so the sum is $n \cdot n^k = n^{k+1}$. This gives the minimum value of the sum, so $C = n^{k+1}$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = n^{k+1}$.
P328. Let $a, b, c, d > 0$ such that $a+b+c=1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c, d$: \(a^{3}+b^{3}+c^{3}+a b c d \geq C.\)
S328. $C = \min \left{\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right}$
Suppose the inequality is false. Then we have, taking into account that $a b c \leq \frac{1}{27}$, the inequality $d\left(\frac{1}{27}-a b c\right)>a^{3}+b^{3}+c^{3}-\frac{1}{9}$. We may assume that $a b c<\frac{1}{27}$. Now, we will reach a contradiction proving that $a^{3}+b^{3}+c^{3}+a b c d \geq \frac{1}{4}$. It is sufficient to prove that\n\n\(\n\\frac{a^{3}+b^{3}+c^{3}-\\frac{1}{9}}{\\frac{1}{27}-a b c} a b c+a^{3}+b^{3}+c^{3} \\geq \\frac{1}{4}\n\)\n\nBut this inequality is equivalent to $4 \sum a^{3}+15 a b c \geq 1$. We use now the identity $\sum a^{3}=3 a b c+1-3 \sum a b$ and reduce the problem to proving that $\sum a b \leq \frac{1+9 a b c}{4}$, which is Schur’s Inequality.\n\nEquality in the above inequalities holds when $a = b = c = \frac{1}{3}$ and $d$ is arbitrary, since then $a^3 + b^3 + c^3 + a b c d = 3 \left(\frac{1}{3}\right)^3 + \left(\frac{1}{27}\right) d = \frac{1}{9} + \frac{d}{27}$. The minimum value of $a^3 + b^3 + c^3 + a b c d$ is also achieved when one of $a, b, c$ is $1$ and the others are $0$, giving $1^3 = 1$. However, since $a, b, c > 0$ and $a + b + c = 1$, the minimum is $\frac{1}{4}$ when two variables approach $0$ and one approaches $\frac{1}{2}$, i.e., $a = b = \frac{1}{2}, c \to 0$. Thus, the minimum value of $C$ is $\min \left\{\frac{1}{4}, \frac{1}{9} + \frac{d}{27}\right\}$.\n\nTherefore, the answer is $C = \min \left\{\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right\}$.
P329. Let $x, y, z > 0$ be real numbers such that $x + y + z = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $x, y, z$: \(\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right) \leq C.\)
S329. $C = \frac{1}{32}$
Let $p=x+y+z=1, q=x y+y z+z x, r=x y z$.\nThen we have\n\(\n\\begin{aligned}\nx^{2}+y^{2} & =(x+y)^{2}-2 x y=(1-z)^{2}-2 x y=1-2 z+z^{2}-2 x y \\\\\n& =1-z-z(1-z)-2 x y=1-z-z(x+y)-2 x y=1-z-q-x y\n\\end{aligned}\n\)\n\nAnalogously we deduce\n\(\ny^{2}+z^{2}=1-x-q-y z \\quad \\text { and } \\quad z^{2}+x^{2}=1-y-q-z x\n\)\n\nSo the given inequality becomes\n\(\n\\begin{equation*}\n(1-z-q-x y)(1-x-q-y z)(1-y-q-z x) \\leq \\frac{1}{32} \\tag{1}\n\\end{equation*}\n\)\n\nAfter algebraic transformations we find that inequality (1) is equivalent to\n\(\n\\begin{equation*}\nq^{2}-2 q^{3}-r(2+r-4 q) \\leq \\frac{1}{32} \\tag{2}\n\\end{equation*}\n\)\n\nAssume that $q \leq \frac{1}{4}$.\nUsing $ p^{3}-4 p q+9 r \geq 0$ , it follows that\n\(\n9 r \\geq 4 q-1, \\quad \\text { i.e. } \\quad r \\geq \\frac{4 q-1}{9}\n\)\nand clearly $q \leq \frac{1}{3}$.\nIt follows that\n\(\n2+r-4 q \\geq 2+\\frac{4 q-1}{9}-4 q=\\frac{17-32 q}{9} \\geq \\frac{17-\\frac{32}{3}}{9}>0\n\)\n\nSo we have\n\(\n\\begin{aligned}\nq^{2}-2 q^{3}-r(2+r-4 q) & \\leq q^{2}-2 q^{3}=q^{2}(1-2 q) \\\\\n& =\\frac{q}{2} \\cdot 2 q(1-2 q) \\leq \\frac{q}{2}\\left(\\frac{2 q+(1-2 q)}{2}\\right)^{2}=\\frac{q}{8} \\leq \\frac{1}{32}\n\\end{aligned}\n\)\ni.e. inequality (2) holds for $q \leq \frac{1}{4}$.\n\nWe need just to consider the case when $q>\frac{1}{4}$.\nLet\n\(\n\\begin{equation*}\nf(r)=q^{2}-2 q^{3}-r(2+r-4 q) \\tag{3}\n\\end{equation*}\n\)\n\nClearly $r \geq \frac{4 q-1}{9}$.\nUsing $p q-9 r \geq 0$ it follows that $9 r \leq q$, i.e. $r \leq \frac{q}{9}$.\nWe have\n\(\nf^{\\prime}(r)=4 q-2-2 r \\leq \\frac{4}{3}-2-2 r \\leq 0\n\)\n\nThis means that $f$ is a strictly decreasing function on $\left(\frac{4 q-1}{9}, \frac{q}{9}\right)$, from which it follows that\n\(\nf(r) \\leq f\\left(\\frac{4 q-1}{9}\\right)=q^{2}-2 q^{3}-\\frac{1}{81}(4 q-1)(17-32 q)\n\)\ni.e.\n\(\n\\begin{equation*}\nf(r) \\leq \\frac{81\\left(q^{2}-2 q^{3}\\right)-(4 q-1)(17-32 q)}{81} \\tag{4}\n\\end{equation*}\n\)\n\nLet\n\(\n\\begin{equation*}\ng(q)=81\\left(q^{2}-2 q^{3}\\right)-(4 q-1)(17-32 q) \\tag{5}\n\\end{equation*}\n\)\n\nThen\n\(\ng^{\\prime}(q)=-486 q^{2}+418 q-100\n\)\n\nSince $\frac{1}{4}<q \leq \frac{1}{3}$, we get\n\(\ng^{\\prime}(q)=-486 q^{2}+418 q-100<\\frac{-486}{16}+\\frac{418}{3}-100<0\n\)\n\nSo $g$ decreases on $(1 / 4,1 / 3)$, i.e. we have\n\(\n\\begin{equation*}\ng(q)<g\\left(\\frac{1}{4}\\right)=\\frac{81}{32} \\tag{6}\n\\end{equation*}\n\)\n\nFinally by (3), (4), (5) and (6) we obtain\n\(\n\\begin{aligned}\nq^{2}-2 q^{3}-r(2+r-4 q) & =f(r) \\leq f\\left(\\frac{4 q-1}{9}\\right) \\\\\n& =\\frac{81\\left(q^{2}-2 q^{3}\\right)-(4 q-1)(17-32 q)}{81} \\\\\n& =\\frac{g(q)}{81}<\\frac{g\\left(\\frac{1}{4}\\right)}{81}=\\frac{\\frac{81}{32}}{81}=\\frac{1}{32}\n\\end{aligned}\n\)\nas required.\n\nEquality holds when $x = y = z = \frac{1}{3}$, since then $x + y + z = 1$ and $\left(x^2 + y^2\right)\left(y^2 + z^2\right)\left(z^2 + x^2\right) = \left(2\left(\frac{1}{3}\right)^2\right)^3 = \left(\frac{2}{9}\right)^3 = \frac{8}{729} = \frac{1}{32}$. This gives the maximum value of the expression, so $C = \frac{1}{32}$ is minimal such that the inequality always holds.\n\nTherefore, the answer is $C = \frac{1}{32}$.
P330. Let $a, b, c$ be real numbers such that $a^3 + b^3 + c^3 - 3abc = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(a^2 + b^2 + c^2 \geq C\)
S330. $C = 1$
Observe that\n\(\n\\begin{aligned}\n1 & =a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right) \\\\\n& =\\frac{(a+b+c)}{2}\\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\\right)\n\\end{aligned}\n\)\n\nSince $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$ we must have $a+b+c>0$.\nAccording to\n\(\n(a+b+c)\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)=1\n\)\nwe deduce\n\(\n(a+b+c)\\left(a^{2}+b^{2}+c^{2}-\\frac{(a+b+c)^{2}-a^{2}-b^{2}-c^{2}}{2}\\right)=1\n\)\nand easily find\n\(\na^{2}+b^{2}+c^{2}=\\frac{1}{3}\\left((a+b+c)^{2}+\\frac{2}{a+b+c}\\right)\n\)\n\nSince $a+b+c>0$ we may use $A M \geq G M$ as follows\n\(\na^{2}+b^{2}+c^{2}=\\frac{1}{3}\\left((a+b+c)^{2}+\\frac{1}{a+b+c}+\\frac{1}{a+b+c}\\right) \\geq 1\n\)\nas required.\nEquality occurs iff $a+b+c=1$.\n\nEquality holds when $a+b+c=1$, which is possible for example when $a = b = c = \frac{1}{3}$. In this case, $a^2 + b^2 + c^2 = 3 \times \left(\frac{1}{3}\right)^2 = 1$. Thus, the minimum value of $a^2 + b^2 + c^2$ is $1$, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.
P331. 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 $a, b, c$: \(a b c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq C.\)
S331. $C = \frac{244}{27}$
By the inequality $A M \geq G M$ we get\n\(\na b c+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq 4 \\sqrt[4]{a b c \\cdot \\frac{1}{a} \\cdot \\frac{1}{b} \\cdot \\frac{1}{c}}=4\n\)\nwith equality if and only if $a b c=\frac{1}{a}=\frac{1}{b}=\frac{1}{c}$, from which we easily deduce that $a=b=c=1$ and then $a+b+c=3$, a contradiction since $a+b+c=1$.\n\nSince $a b c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is symmetrical with respect to $a, b$ and $c$ we estimate that the minimal value occurs when $a=b=c$, i.e. $a=b=c=1 / 3$, since $a+b+c=1$.\n\nLet $a b c=\frac{1}{\alpha a}=\frac{1}{\alpha b}=\frac{1}{\alpha c}$, from which we obtain $\alpha=\frac{1}{a^{2} b c}=81$.\n\nTherefore let us rewrite the given expression as follows\n\(\n\\begin{equation*}\na b c+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=a b c+\\frac{1}{81 a}+\\frac{1}{81 b}+\\frac{1}{81 c}+\\frac{80}{81}\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\tag{1}\n\\end{equation*}\n\)\n\nBy $A M \geq G M$ and $A M \geq H M$ we have\n\(\n\\begin{equation*}\na b c+\\frac{1}{81 a}+\\frac{1}{81 b}+\\frac{1}{81 c} \\geq 4 \\sqrt[4]{a b c \\cdot \\frac{1}{81 a} \\cdot \\frac{1}{81 b} \\cdot \\frac{1}{81 c}}=\\frac{4}{27} \\tag{2}\n\\end{equation*}\n\)\nand\n\(\n\\begin{equation*}\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq \\frac{9}{a+b+c}=9 \\tag{3}\n\\end{equation*}\n\)\n\nBy (1), (2) and (3) we have\n\(\na b c+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq \\frac{4}{27}+\\frac{80}{9}=\\frac{244}{27}\n\)\nwith equality if and only if $a=b=c=\frac{1}{3}$.\n\nEquality is achieved when $a = b = c = \frac{1}{3}$, which satisfies $a + b + c = 1$. This gives the minimum value of the expression, so $C = \frac{244}{27}$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{244}{27}$.
P332. Let $a, b, c \in \mathbb{R}^{+}$ such that $a + 2b + 3c \geq 20$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: \(S = a + b + c + \frac{3}{a} + \frac{9}{2b} + \frac{4}{c} \geq C\)
S332. $C = 13$
$S=13$ at the point $a=2, b=3, c=4$.\nUsing $A M \geq G M$ we get\n\(\na+\\frac{4}{a} \\geq 2 \\sqrt{a \\cdot \\frac{4}{a}}=4, \\quad b+\\frac{9}{b} \\geq 2 \\sqrt{b \\cdot \\frac{9}{b}}=6, \\quad c+\\frac{16}{c} \\geq 2 \\sqrt{c \\cdot \\frac{16}{c}}=8\n\)\ni.e.\n\(\n\\frac{3}{4}\\left(a+\\frac{4}{a}\\right) \\geq 3, \\quad \\frac{1}{2}\\left(b+\\frac{9}{b}\\right) \\geq 3 \\quad \\text { and } \\quad \\frac{1}{4}\\left(c+\\frac{16}{c}\\right) \\geq 2\n\)\n\nAdding the last three inequalities we have\n\(\n\\begin{equation*}\n\\frac{3}{4} a+\\frac{1}{2} b+\\frac{1}{4} c+\\frac{3}{a}+\\frac{9}{2 b}+\\frac{4}{c} \\geq 8 \\tag{1}\n\\end{equation*}\n\)\n\nUsing $a+2 b+3 c \geq 20$ we obtain\n\(\n\\begin{equation*}\n\\frac{1}{4} a+\\frac{1}{2} b+\\frac{3}{4} c \\geq 5 \\tag{2}\n\\end{equation*}\n\)\n\nFinally, after adding (1) and (2) we get\n\(\na+b+c+\\frac{3}{a}+\\frac{9}{2 b}+\\frac{4}{c} \\geq 13\n\)\nas desired.\n\nEquality holds when $a=2$, $b=3$, $c=4$, since at this point $a+2b+3c=2+6+12=20$ and $S=2+3+4+\frac{3}{2}+\frac{9}{6}+1=9+1.5+1.5+1=13$. This gives the minimum value of $S$, so $C=13$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 13$.
P333. Let $a, b, c, d > 0$ be real numbers. Let $u = ab + ac + ad + bc + bd + cd$ and $v = abc + abd + acd + bcd$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c, d > 0$: \(2u^3 \geq Cv^2\)
S333. $C = 27$
We have $p_{2}=\frac{u}{\binom{4}{2}}=\frac{u}{6}$ and $p_{3}=\frac{v}{\binom{4}{3}}=\frac{v}{4}$.\nBy Maclaurin’s inequality we have\n\(\np_{2}^{\\frac{1}{2}} \\geq p_{3}^{\\frac{1}{3}} \\quad \\Leftrightarrow \\quad p_{2}^{3} \\geq p_{3}^{2} \\quad \\Leftrightarrow \\quad\\left(\\frac{u}{6}\\right)^{3} \\geq\\left(\\frac{v}{4}\\right)^{2} \\quad \\Leftrightarrow \\quad 2 u^{3} \\geq 27 v^{2}\n\)\n\nEquality holds when $a = b = c = d$, that is, all variables are equal. In this case, the inequality becomes an equality, and $C = 27$ is the maximal value for which the inequality always holds.\n\nTherefore, the answer is $C = 27$.
P334. Let $x, y, z \in \mathbb{R}^{+}$ such that $x + y + z = 1$. Find the maximal constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint: \(\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} \geq C\)
S334. $C = 1$
We have\n\(\n\\begin{equation*}\n\\frac{x y}{z}+\\frac{y z}{x}+\\frac{z x}{y}=\\frac{1}{2}\\left(\\frac{x y}{z}+\\frac{y z}{x}\\right)+\\frac{1}{2}\\left(\\frac{y z}{x}+\\frac{z x}{y}\\right)+\\frac{1}{2}\\left(\\frac{z x}{y}+\\frac{x y}{z}\\right) \\tag{1}\n\\end{equation*}\n\)\n\nSince $A M \geq G M$ we have\n\(\n\\frac{1}{2}\\left(\\frac{x y}{z}+\\frac{y z}{x}\\right) \\geq \\sqrt{\\frac{x y}{z} \\frac{y z}{x}}=y\n\)\n\nAnalogously we get\n\(\n\\frac{1}{2}\\left(\\frac{y z}{x}+\\frac{z x}{y}\\right) \\geq z \\quad \\text { and } \\quad \\frac{1}{2}\\left(\\frac{z x}{y}+\\frac{x y}{z}\\right) \\geq x\n\)\n\nAdding these three inequalities we obtain\n\(\n\\frac{x y}{z}+\\frac{y z}{x}+\\frac{z x}{y} \\geq x+y+z=1\n\)\n\nEquality holds if and only if $x = y = z$, that is, when $x = y = z = \frac{1}{3}$. In this case, the minimum value of the expression is achieved, so $C = 1$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.
P335. Let $a, b, c, d$ be non-negative real numbers. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c, d \in \mathbb{R}^{+}$: \(a^{4}+b^{4}+c^{4}+d^{4}+C a b c d \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}+a^{2} c^{2}+b^{2} d^{2}\)
S335. $C = 2$
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 $C = 2$.
P336. 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 $\triangle ABC$: \(\frac{L P}{R^3} \leq C.\)
S336. $C = \frac{27}{4}$
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 $C = \frac{27}{4}$.
P337. 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: \(\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+d)}+\frac{1}{d(1+a)} \geq C\)
S337. $C = 2$
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 $C = 2$.
P338. Let $x, y, z$ be real numbers. Find the largest constant $C$ such that the following inequality holds for all $x, y, z \in \mathbb{R}$: \(x^{4}+y^{4}+z^{4} \geq 4 x y z + C\)
S338. $C = -1$
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 $C = -1$.
P339. 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: \(\frac{1+a^{2}}{1-a^{2}}+\frac{1+b^{2}}{1-b^{2}}+\frac{1+c^{2}}{1-c^{2}} \geq C\)
S339. $C = \frac{15}{4}$
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 $C = \frac{15}{4}$.
P340. Let $a, b, c \in \mathbb{R}^{+}$. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{abc}{(1+a)(a+b)(b+c)(c+16)} \leq C.\]S340. $C = \frac{1}{81}$
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 $C = \frac{1}{81}$.
P341. Let $a, b, c \in (1,2)$ be real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in (1,2)$: \(\frac{b \sqrt{a}}{4 b \sqrt{c}-c \sqrt{a}}+\frac{c \sqrt{b}}{4 c \sqrt{a}-a \sqrt{b}}+\frac{a \sqrt{c}}{4 a \sqrt{b}-b \sqrt{c}} \geq C\)
S341. $C = 1$
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 $C = 1$.
P342. 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 $a, b, c$: \(\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geq C\)
S342. $C = \frac{3}{2}$
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 $C = \frac{3}{2}$.
P343. Find the smallest constant $C$ such that for all real numbers $x$, the following inequality holds: \(2 x^{4} + C \geq 2 x^{3} + x^{2}\)
S343. $C = 1$
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 $C = 1$.
P344. 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 $a, b, c$: \(\frac{a^{3}+2}{b+2}+\frac{b^{3}+2}{c+2}+\frac{c^{3}+2}{a+2} \geq C.\)
S344. $C = 3$
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 $C = 3$.
P345. 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: \(\frac{l_{\alpha}}{a}+\frac{l_{\beta}}{b}+\frac{l_{\gamma}}{c} \leq C\frac{s}{r}.\)
S345. $C = \frac{1}{2}$
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 $C = \frac{1}{2}$.
P346. 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: \(a^{2}+b^{2}+c^{2} < C(a b+b c+c a)\)
S346. $C = 2$
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 $C = 2$.
P347. 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: \(\sqrt[3]{a b + b c} + \sqrt[3]{b c + c a} + \sqrt[3]{c a + a b} \leq C\)
S347. $C = 6$
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 $C = 6$.
P348. 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: \(\sqrt[n]{x_{1} x_{2} \cdots x_{n}} \geq C\)
S348. $C = 1998(n-1)$
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 $C = 1998(n-1)$.
P349. 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: \(\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq C\)
S349. $C = 1$
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 $C = 1$.
P350. 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: \(\left(\frac{3-x}{1-x}\right)^{2}+\left(\frac{3-y}{1-y}\right)^{2}+\left(\frac{3-z}{1-z}\right)^{2} > C\)
S350. $C = 7$
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 $C = 7$.
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