Inequality Proof Problems [301-350]
Recommended posting: 【Algebra】 Algebra Index
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
\[\begin{aligned} \left(a^{b} b^{c} c^{a}\right)^{\frac{1}{a+b+c}} &= a^{\frac{b}{a+b+c}}\cdot b^{\frac{c}{a+b+c}}\cdot c^{\frac{a}{a+b+c}} \\ &\le \frac{ba+cb+ac}{a+b+c} \le \frac{(a+b+c)^2}{3(a+b+c)} = \frac{a+b+c}{3}. \end{aligned}\]Equality holds when \(a=b=c\), in which case
\[a^b b^c c^a = a^{a}a^{a}a^{a}=a^{3a} \quad\text{and}\quad \left(C(a+b+c)\right)^{a+b+c}=(3Ca)^{3a}.\]So
\[a^{3a}\le (3Ca)^{3a} \quad\Rightarrow\quad 1\le 3C \quad\Rightarrow\quad C\ge \frac{1}{3}.\]Thus, the minimal value of \(C\) is \(\frac{1}{3}\).
Therefore, 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
\[a^{2}b+b^{2}c+c^{2}a+3abc \ge 2\left(ab^{2}+bc^{2}+ca^{2}\right)\]with equality if and only if \(a=b=c\), i.e. the triangle is equilateral.
Let us use Ravi’s substitutions, i.e.
\[a=x+y,\quad b=y+z,\quad c=z+x.\]Then the given inequality becomes
\[x^{3}+y^{3}+z^{3}+x^{2}y+y^{2}z+z^{2}x \ge 2\left(x^{2}z+y^{2}x+z^{2}y\right).\]Since \(AM \ge GM\) we have
\[x^{3}+z^{2}x \ge 2x^{2}z,\quad y^{3}+x^{2}y \ge 2y^{2}x,\quad z^{3}+y^{2}z \ge 2z^{2}y.\]After adding these inequalities we obtain
\[x^{3}+y^{3}+z^{3}+x^{2}y+y^{2}z+z^{2}x \ge 2\left(x^{2}z+y^{2}x+z^{2}y\right).\]Equality 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.
Therefore, 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 \(\overline{EF}=a\cos\alpha\), \(\overline{FD}=b\cos\beta\), \(\overline{DE}=c\cos\gamma\), and the given inequality becomes
\[\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma \ge \frac{3}{4}.\]Equality holds when the triangle is equilateral, i.e., \(\alpha=\beta=\gamma=60^\circ\), so
\[\cos^{2}60^\circ=\left(\frac{1}{2}\right)^2=\frac{1}{4}\]and the sum is \(3\cdot\frac{1}{4}=\frac{3}{4}\). This gives the minimum value, so \(C=\frac{3}{4}\) is the largest constant for which the inequality always holds.
Therefore, 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:
\[(a+b)(b+c)(c+a) \geq C(a+b+c-1)\]S304.1. $C = 4$
We will use the fact that
\[(a+b)(b+c)(c+a)\ge \frac{8}{9}(a+b+c)(ab+bc+ca).\]So, it is enough to prove that
\[\frac{2}{9}(ab+bc+ca)+\frac{1}{a+b+c}\ge 1.\]Using the AM-GM inequality, we can write
\[\frac{2}{9}(ab+bc+ca)+\frac{1}{a+b+c} \ge 3\sqrt[3]{\frac{(ab+bc+ca)^2}{81(a+b+c)}}\ge 1,\]because
\[(ab+bc+ca)^2 \ge 3abc(a+b+c)=3(a+b+c).\]Equality 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.
Therefore, the answer is \(C=4\).
S304.2.
Using the identity
\[(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc\]we reduce the problem to the following one
\[ab+bc+ca+\frac{3}{a+b+c}\ge 4.\]Now, we can apply the AM-GM inequality in the following form
\[ab+bc+ca+\frac{3}{a+b+c} \ge 4\sqrt[4]{\frac{(ab+bc+ca)^3}{(a+b+c)}}.\]And so it is enough to prove that
\[(ab+bc+ca)^3 \ge (a+b+c).\]But this is easy, because we clearly have \(ab+bc+ca\ge 3\) and
\[(ab+bc+ca)^2 \ge 3abc(a+b+c)=3(a+b+c).\]Equality holds when \(a=b=c=1\), since then \(abc=1\), \(a+b+c=3\), \(ab+bc+ca=3\), and \((a+b)(b+c)(c+a)=8\). This gives the minimum value of \((a+b)(b+c)(c+a)\) for the given constraint, so \(C=4\) is the maximal constant.
Therefore, 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 \(x=\frac{1}{bc},\ y=\frac{1}{ac},\ z=\frac{1}{ab}\) and \(A=ac,\ B=ab,\ C=bc\). We have
\[\begin{aligned} \frac{x}{y+z}(B+C) &= ab\frac{a+c}{b+c},\\ \frac{y}{z+x}(C+A) &= bc\frac{b+a}{c+a},\\ \frac{z}{x+y}(A+B) &= ca\frac{c+b}{a+b}. \end{aligned}\]Using the following corollary: let \(a,b,c\) and \(x,y,z\) be positive real numbers. Then
\[\frac{x}{y+z}(b+c)+\frac{y}{z+x}(c+a)+\frac{z}{x+y}(a+b)\ge \sqrt{3(ab+bc+ca)},\]and the previous identities we obtain
\[\begin{aligned} &ab\frac{a+c}{b+c}+bc\frac{b+a}{c+a}+ca\frac{c+b}{a+b}\\ &\quad=\frac{x}{y+z}(B+C)+\frac{y}{z+x}(C+A)+\frac{z}{x+y}(A+B)\\ &\quad\ge \sqrt{3(AB+BC+CA)}=\sqrt{3abc(a+b+c)}. \end{aligned}\]Equality 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.
Therefore, 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 \(w=ax+by\). Then
\[\begin{aligned} uv&=(a^{2}+b^{2})(x^{2}+y^{2})\\ &=a^{2}x^{2}+a^{2}y^{2}+b^{2}x^{2}+b^{2}y^{2}\\ &=a^{2}x^{2}+b^{2}y^{2}+2axby+a^{2}y^{2}+b^{2}x^{2}-2axby\\ &=(ax+by)^{2}+(ay-bx)^{2}\\ &=w^{2}+1. \end{aligned}\]From the obvious inequality \((t\sqrt{3}+1)^{2}\ge 0\) we deduce
\[3t^{2}+1\ge -2t\sqrt{3},\]i.e.
\[4t^{2}+4\ge 3-2t\sqrt{3}+t^{2},\]i.e.
\[4t^{2}+4\ge (\sqrt{3}-t)^{2}. \qquad (1)\]Now we have
\[(u+v)^{2}\ge 4uv=4(w^{2}+1)\stackrel{(1)}{\ge}(\sqrt{3}-w)^{2},\]from which we get \(u+v\ge \sqrt{3}-w\), which is equivalent to
\[u+v+w\ge \sqrt{3}.\]Equality holds when \(ay-bx=1\) and \(ax+by=-\frac{\sqrt{3}}{2}\), with
\[a^{2}+b^{2}=x^{2}+y^{2}=\frac{\sqrt{3}}{2}.\]This gives the minimum value of \(a^{2}+b^{2}+x^{2}+y^{2}+ax+by\), so \(C=\sqrt{3}\) is the largest constant for which the inequality always holds.
Therefore, 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 among the numbers \(x,y,z\), then \(z\ge \frac{1}{2}\); we saw that
\[\begin{aligned} \frac{1}{x}+\frac{1}{y}-4(x+y) &=(x+y)\left(\frac{1}{xy}-4\right)\\ &\ge \frac{2}{2z+1}(2z+1)\\ &=\frac{2(2z-1)(2z+3)}{2z+1} =4z-\frac{1}{z}+\frac{(2z-1)^{2}}{z(2z+1)}. \end{aligned}\]from where we get the inequality
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-4(x+y+z)\ge \frac{(2z-1)^{2}}{z(2z+1)}.\]Of course, in the right-hand side \(z\) could be replaced by any of the three numbers which is \(\ge \frac{1}{2}\) (surely there is at least one such number).
Equality 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 \(C=4\).
Therefore, 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 \(a=xy,\ b=yz,\ c=zx\). Then \(xy+yz+zx=1\) and we may take
\[x=\tan\frac{\alpha}{2},\quad y=\tan\frac{\beta}{2},\quad z=\tan\frac{\gamma}{2}\]where \(\alpha,\beta,\gamma\in(0,\pi)\) and \(\alpha+\beta+\gamma=\pi\).
We have
\[\begin{aligned} \sqrt{\frac{1}{a}-1}\sqrt{\frac{1}{b}-1} &=\sqrt{\frac{(1-a)(1-b)}{ab}} =\sqrt{\frac{(1-xy)(1-yz)}{xy^{2}z}}\\ &=\sqrt{\frac{(yz+zx)(zx+xy)}{xy^{2}z}} =\sqrt{\frac{z(x+y)\cdot x(y+z)}{xy^{2}z}}\\ &=\sqrt{\frac{(x+y)(y+z)}{y^{2}}} =\frac{\sqrt{(1+y^{2})}}{y}\\ &=\frac{\sqrt{1+\tan^{2}\frac{\beta}{2}}}{\tan\frac{\beta}{2}} =\frac{1}{\sin\frac{\beta}{2}}. \end{aligned}\]Similarly we obtain
\[\sqrt{\frac{1}{b}-1}\sqrt{\frac{1}{c}-1}=\frac{1}{\sin\frac{\gamma}{2}} \quad\text{and}\quad \sqrt{\frac{1}{c}-1}\sqrt{\frac{1}{a}-1}=\frac{1}{\sin\frac{\alpha}{2}}.\]Now the given inequality becomes
\[\frac{1}{\sin\frac{\alpha}{2}}+\frac{1}{\sin\frac{\beta}{2}}+\frac{1}{\sin\frac{\gamma}{2}}\ge 6.\]By \(AM\ge HM\) we have
\[\frac{1}{\sin\frac{\alpha}{2}}+\frac{1}{\sin\frac{\beta}{2}}+\frac{1}{\sin\frac{\gamma}{2}} \ge \frac{9}{\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}}.\]So we need to prove that
\[\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}\le \frac{3}{2}.\]Equality occurs if and only if \(\alpha=\beta=\gamma=\frac{\pi}{3}\), 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.
Therefore, 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 \(xyzt=1\) and so the problem becomes:
If \(a,b,c,d\) have product \(1\), then
\[a^{2}+b^{2}+c^{2}+d^{2}+2 \ge ab+bc+cd+da+ac+bd.\]Let \(d\) be the minimum among \(a,b,c,d\) and let \(m=\sqrt[3]{abc}\). We will prove that
\[a^{2}+b^{2}+c^{2}+d^{2}+2-(ab+bc+cd+da+ac+bd) \ge d^{2}+3m^{2}+2-\left(3m^{2}+3md\right),\]which is in fact
\[a^{2}+b^{2}+c^{2}-ab-bc-ca \ge d\left(a+b+c-3\sqrt[3]{abc}\right).\]Because \(d\le \sqrt[3]{abc}\), proving this first inequality comes down to the inequality
\[a^{2}+b^{2}+c^{2}-ab-bc-ca \ge \sqrt[3]{abc}\left(a+b+c-3\sqrt[3]{abc}\right).\]Take
\[u=\frac{a}{\sqrt[3]{abc}},\quad v=\frac{b}{\sqrt[3]{abc}},\quad w=\frac{c}{\sqrt[3]{abc}}.\]Using Problem 74, we find that
\[u^{2}+v^{2}+w^{2}+3 \ge u+v+w+uv+vw+wu,\]which is exactly
\[a^{2}+b^{2}+c^{2}-ab-bc-ca \ge \sqrt[3]{abc}\left(a+b+c-3\sqrt[3]{abc}\right).\]Thus, it remains to prove that
\[d^{2}+2 \ge 3md \Longleftrightarrow d^{2}+2 \ge 3\sqrt[3]{d^{2}},\]which is clear.
Equality 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.
Therefore, 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:
\[\begin{gathered} mn(x-y)\left(x^{m+n-1}-y^{m+n-1}\right) \ge (m+n-1)\left(x^{m}-y^{m}\right)\left(x^{n}-y^{n}\right) \ \Longleftrightarrow \\ \Longleftrightarrow \frac{x^{m+n-1}-y^{m+n-1}}{(m+n-1)(x-y)} \ge \frac{x^{m}-y^{m}}{m(x-y)}\cdot \frac{x^{n}-y^{n}}{n(x-y)}. \end{gathered}\](we have assumed that \(x>y\)). The last relation can also be written
\[(x-y)\int_{y}^{x} t^{m+n-2}\,dt \ge \left(\int_{y}^{x} t^{m-1}\,dt\right)\left(\int_{y}^{x} t^{n-1}\,dt\right),\]and this follows from Chebyshev’s inequality for integrals.
Equality 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 \(C=(n-1)(m-1)\).
Therefore, 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
\[\left(a^{a}+b^{a}+c^{a}\right)^{\frac{1}{3}} \left(a^{b}+b^{b}+c^{b}\right)^{\frac{1}{3}} \left(a^{c}+b^{c}+c^{c}\right)^{\frac{1}{3}} \ge a^{\frac{a+b+c}{3}}+b^{\frac{a+b+c}{3}}+c^{\frac{a+b+c}{3}}.\]Since \(a+b+c=1\), the conclusion follows.
Equality 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.
Therefore, 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 \(y=ab+bc+ca\). Using brute-force, it is easy to see that the left hand side is
\[\frac{x^{2}+4x+y+3}{x^{2}+2x+y+xy},\]while the right hand side is
\[\frac{12+4x+y}{9+4x+2y}.\]Now, the inequality becomes
\[\frac{x^{2}+4x+y+3}{x^{2}+2x+y+xy}-1 \le \frac{12+4x+y}{9+4x+2y}-1 \ \Longleftrightarrow\ \frac{2x+3-xy}{x^{2}+2x+y+xy} \le \frac{3-y}{9+4x+2y}.\]For the last inequality, we clear denominators. Then using the inequalities
\[x\ge 3,\quad y\ge 3,\quad x^{2}\ge 3y,\]we have
\[\frac{5}{3}x^{2}y \ge 5x^{2},\quad \frac{x^{2}y}{3}\ge y^{2},\quad xy^{2}\ge 9x,\quad 5xy\ge 15x,\quad xy\ge 3y,\quad x^{2}y\ge 27.\]Summing up these inequalities, the desired inequality follows.
Equality 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.
Therefore, 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 \(x=a+b+c=3\), \(y=ab+bc+ca\), \(z=abc\). Now the given inequality can be rewritten as
\[z^{2}-2z-2xz+z(x+y)+x^{2}+x+y^{2}-y+3xy+1 \ge 9y,\]i.e.
\[(z-1)^{2}-(z-1)(x-y)+(x-y)^{2}\ge 0,\]which is obviously true. Equality holds iff \(a=b=c=1\).
Equality 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 \(27 \ge 9\cdot 3=27\). Thus, \(C=9\) is the largest constant for which the inequality always holds, and this gives the minimum value of \(C\).
Therefore, 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 \(1-a^{2}>0\), \(1-b^{2}>0\), \(1-c^{2}>0\). By \(AM\ge GM\) we get
\[\begin{aligned} 6\sqrt[3]{(1-a^{2})(1-b^{2})(1-c^{2})} &=2\cdot 3\sqrt[3]{(1-a^{2})(1-b^{2})(1-c^{2})}\\ &\le 2\left((1-a^{2})+(1-b^{2})+(1-c^{2})\right)\\ &=2\left(3-(a^{2}+b^{2}+c^{2})\right)\\ &=6-2(a^{2}+b^{2}+c^{2}). \end{aligned}\]We’ll show that
\[6-2(a^{2}+b^{2}+c^{2}) \le 1+(a+b+c)^{2}.\]This inequality is equivalent to
\[6-2(a^{2}+b^{2}+c^{2}) \le 1+a^{2}+b^{2}+c^{2}+2(ab+bc+ca),\]i.e.
\[3 \le 3(a^{2}+b^{2}+c^{2})+2(ab+bc+ca).\]Since \(ab+bc+ca=1\), it is enough to prove that
\[a^{2}+b^{2}+c^{2}\ge 1,\]which is true because
\[a^{2}+b^{2}+c^{2}\ge ab+bc+ca=1.\]Equality 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.
Therefore, 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:
\[xy + xz + yz \geq C + \sqrt{x^2 + 1} + \sqrt{y^2 + 1} + \sqrt{z^2 + 1}\]S315.1. $C = 3$
Another improvement is as follows. Start from
\[\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}} \ge \frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}=1 \ \Rightarrow\ x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}\ge x^{2}y^{2}z^{2}.\]which is equivalent to
\[(xy+xz+yz)^{2}\ge 2xyz(x+y+z)+x^{2}y^{2}z^{2}=3(x+y+z)^{2}.\]Further on,
\[\begin{aligned} (xy+xz+yz-3)^{2} &=(xy+xz+yz)^{2}-6(xy+xz+yz)+9\\ &\ge 3(x+y+z)^{2}-6(xy+xz+yz)+9\\ &=3(x^{2}+y^{2}+z^{2})+9, \end{aligned}\]so that
\[xy+xz+yz \ge 3+\sqrt{3(x^{2}+y^{2}+z^{2})+9}.\]But
\[\sqrt{3(x^{2}+y^{2}+z^{2})+9}\ge \sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}\]is a consequence of the Cauchy-Schwarz inequality and we have a second improvement and proof for the desired inequality:
\[\begin{aligned} xy+xz+yz &\ge 3+\sqrt{3(x^{2}+y^{2}+z^{2})+9}\\ &\ge 3+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}. \end{aligned}\]Equality 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,
\[xy+xz+yz=3\cdot 3=9 \quad\text{and}\quad \sqrt{x^{2}+1}=\sqrt{3+1}=2\]for each variable, so \(C=9-3\cdot 2=3\). This gives the minimum value of
\[xy+xz+yz-\left(\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}\right),\]so the maximal constant \(C\) is \(3\).
Therefore, the answer is \(C=3\).
S315.2. $C = 3$
We have
\[xyz=x+y+z \ge 2\sqrt{xy}+z \ \Rightarrow\ z(\sqrt{xy})^{2}-2\sqrt{xy}-z\ge 0.\]Because the positive root of the trinomial \(zt^{2}-2t-z\) is
\[\frac{1+\sqrt{1+z^{2}}}{z},\]we get from here
\[\sqrt{xy}\ge \frac{1+\sqrt{1+z^{2}}}{z} \ \Longleftrightarrow\ z\sqrt{xy}\ge 1+\sqrt{1+z^{2}}.\]Of course, we have two other similar inequalities. Then,
\[\begin{aligned} xy+xz+yz &\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}\\ &\ge 3+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}, \end{aligned}\]and we have both a proof of the given inequality, and a little improvement of it.
Equality holds when \(x=y=z=\sqrt{3}\), since then \(x+y+z=3\sqrt{3}=xyz\), and
\[xy+xz+yz=9,\quad \sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}=6,\]so the difference equals \(3\), i.e. \(C=3\).
Therefore, 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
\[\begin{aligned} &\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\left(a(b+2c)+b(c+2a)+c(a+2b)\right)\\ &\quad\ge (a+b+c)^{2}. \end{aligned}\]hence
\[\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \ge \frac{(a+b+c)^{2}}{a(b+2c)+b(c+2a)+c(a+2b)} = \frac{(a+b+c)^{2}}{3(ab+bc+ca)}.\]So it suffices to show that
\[\frac{(a+b+c)^{2}}{3(ab+bc+ca)}\ge 1, \quad\text{i.e.}\quad (a+b+c)^{2}\ge 3(ab+bc+ca),\]which is equivalent to \(a^{2}+b^{2}+c^{2}\ge ab+bc+ca\), and clearly holds. Equality occurs iff \(a=b=c\).
Equality 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.
Therefore, 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 \(c=z+x\). So we have
\[x+y+z=\frac{3}{2},\]and since \(AM\ge GM\) we get
\[xyz\le \left(\frac{x+y+z}{3}\right)^{3}=\frac{1}{8}.\]Now we obtain
\[\begin{aligned} a^{2}+b^{2}+c^{2}+\frac{4abc}{3} &=\frac{(a^{2}+b^{2}+c^{2})(a+b+c)+4abc}{3}\\ &=\frac{2\left((x+y)^{2}+(y+z)^{2}+(z+x)^{2}\right)(x+y+z)+4(x+y)(y+z)(z+x)}{3}\\ &=\frac{4}{3}\left((x+y+z)^{3}-xyz\right)\\ &\ge \frac{4}{3}\left(\left(\frac{3}{2}\right)^{3}-\frac{1}{8}\right)=\frac{13}{3}. \end{aligned}\]Equality 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.
Therefore, 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
\[\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)^{2}\ge 9 \ \Longleftrightarrow\ \frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}}+2(a^{2}+b^{2}+c^{2})\ge 3(a^{2}+b^{2}+c^{2}),\]i.e.
\[\frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}} \ge a^{2}+b^{2}+c^{2}.\]Furthermore, applying \(AM\ge GM\) we get
\[\frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}} \ge 2b^{2},\quad \frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}} \ge 2c^{2},\quad \frac{a^{2}b^{2}}{c^{2}}+\frac{c^{2}a^{2}}{b^{2}} \ge 2a^{2}.\]After adding these inequalities we obtain
\[\frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}} \ge a^{2}+b^{2}+c^{2},\]and we are done.
Equality 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 \(C=3\).
Therefore, 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
\[\frac{xy}{xy+(yz)(zx)}+\frac{yz}{yz+(zx)(xy)}+\frac{zx}{zx+(xy)(yz)} \le 1+\frac{3\sqrt{3}}{4},\]i.e.
\[\frac{1}{1+z^{2}}+\frac{1}{1+x^{2}}+\frac{y}{1+y^{2}} \le 1+\frac{3\sqrt{3}}{4}, \qquad (1)\]where \(xy+yz+zx=1\).
Since \(xy+yz+zx=1\), we may set
\[x=\tan\frac{\alpha}{2},\quad y=\tan\frac{\beta}{2},\quad z=\tan\frac{\gamma}{2},\]where \(\alpha,\beta,\gamma\in(0,\pi)\) and \(\alpha+\beta+\gamma=\pi\).
Then inequality (1) becomes
\[\frac{1}{1+\tan^{2}\frac{\gamma}{2}}+\frac{1}{1+\tan^{2}\frac{\alpha}{2}} +\frac{\tan\frac{\beta}{2}}{1+\tan^{2}\frac{\beta}{2}} \le 1+\frac{3\sqrt{3}}{4},\]i.e.
\[\cos^{2}\frac{\gamma}{2}+\cos^{2}\frac{\alpha}{2}+\frac{\sin\beta}{2} \le 1+\frac{3\sqrt{3}}{4}.\]Using the trigonometric identity \(\cos x=2\cos^{2}\frac{x}{2}-1\), the last inequality becomes
\[\frac{\cos\gamma+1}{2}+\frac{\cos\alpha+1}{2}+\frac{\sin\beta}{2} \le 1+\frac{3\sqrt{3}}{4},\]i.e.
\[\cos\gamma+\cos\alpha+\sin\beta \le \frac{3\sqrt{3}}{2}. \qquad (2)\]We have
\[\begin{aligned} \cos\alpha+\cos\gamma+\sin\beta &=\cos\alpha+\cos\gamma+\sin(\pi-(\alpha+\gamma))\\ &=\cos\alpha+\cos\gamma+\sin(\alpha+\gamma)\\ &=\cos\alpha+\cos\gamma+\sin\alpha\cos\gamma+\cos\alpha\sin\gamma\\ &=\frac{2}{\sqrt{3}}\left(\frac{\sqrt{3}}{2}\cos\alpha+\frac{\sqrt{3}}{2}\cos\gamma\right) +\frac{1}{\sqrt{3}}\left(\sqrt{3}\sin\alpha\cos\gamma+\sqrt{3}\cos\alpha\sin\gamma\right)\\ &\le \frac{1}{\sqrt{3}}\left(\frac{3}{4}+\cos^{2}\alpha+\frac{3}{4}+\cos^{2}\gamma\right)\\ &=\frac{1}{2\sqrt{3}}\left(3\sin^{2}\alpha+\cos^{2}\alpha+3\sin^{2}\gamma+\cos^{2}\gamma\right)\\ &=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\left(\sin^{2}\alpha+\cos^{2}\alpha\right) +\frac{\sqrt{3}}{2}\left(\sin^{2}\gamma+\cos^{2}\gamma\right)\\ &=\frac{3\sqrt{3}}{2}. \end{aligned}\]Equality in the above inequalities is achieved when \(\alpha=\beta=\gamma=\frac{\pi}{3}\), 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.
Therefore, 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 $x, y, z \in \mathbb{R}^{+}$:
\[\frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}} \leq C\]S320.1. $C = 1$
From Huygens inequality we have
\[\sqrt{(x+y)(x+z)}\ge x+\sqrt{yz},\]and using this inequality for the similar ones we get
\[\begin{aligned} \frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}} &\le \frac{x}{2x+\sqrt{yz}}+\frac{y}{2y+\sqrt{zx}}+\frac{z}{2z+\sqrt{xy}}. \end{aligned}\]Now, we denote
\[a=\frac{\sqrt{yz}}{x},\quad b=\frac{\sqrt{zx}}{y},\quad c=\frac{\sqrt{xy}}{z},\]and the inequality becomes
\[\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\le 1.\]From the above notations we can see that \(abc=1\), so the last inequality becomes, after clearing denominators,
\[ab+bc+ca\ge 3,\]which follows from the AM-GM inequality.
Equality 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.
Therefore, the answer is \(C=1\).
S320.2. $C = 1$
We have
\[(x+y)(x+z)=xy+\left(x^{2}+yz\right)+xz \ge xy+2x\sqrt{yz}+xz=(\sqrt{xy}+\sqrt{xz})^{2}.\]Hence
\[\sum \frac{x}{x+\sqrt{(x+y)(x+z)}} \le \sum \frac{x}{x+\sqrt{xy}+\sqrt{xz}}.\]But
\[\sum \frac{x}{x+\sqrt{xy}+\sqrt{xz}} =\sum \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} =1,\]and this solves the problem.
Equality holds when \(x=y=z\), since then each term becomes
\[\frac{x}{x+\sqrt{x^{2}}+\sqrt{x^{2}}}=\frac{x}{x+x+x}=\frac{1}{3},\]and the sum is \(1\). This gives the maximum value of the sum, so \(C=1\) is minimal such that the inequality always holds.
Therefore, 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 \(\begin{gathered} \frac{a}{4b^{2}+bc+4c^{2}}+\frac{b}{4c^{2}+ca+4a^{2}}+\frac{c}{4a^{2}+ab+4b^{2}}\\ \ge \frac{(a+b+c)^{2}}{a(4b^{2}+bc+4c^{2})+b(4c^{2}+ca+4a^{2})+c(4a^{2}+ab+4b^{2})}. \end{gathered}\)
So we need to prove that
\[\frac{(a+b+c)^{2}}{4a(b^{2}+c^{2})+4b(c^{2}+a^{2})+4c(a^{2}+b^{2})+3abc} \ge \frac{1}{a+b+c},\]which is equivalent to
\[(a+b+c)^{3}\ge 4a(b^{2}+c^{2})+4b(c^{2}+a^{2})+4c(a^{2}+b^{2})+3abc,\]i.e.
\[a^{3}+b^{3}+c^{3}+3abc \ge a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2}),\]which is Schur’s inequality.
Equality 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.
Therefore, 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 \((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\ge 0\) we deduce
\[2(a^{2}+b^{2}+c^{2}) \ge 2(ab+bc+ca) \ \Longleftrightarrow\ a^{2}+b^{2}+c^{2}\ge ab+bc+ca.\]Equality 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.
Therefore, 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
\[\begin{aligned} A&=\frac{a_{1}}{\sqrt{1-a_{1}}}+\frac{a_{2}}{\sqrt{1-a_{2}}}+\cdots+\frac{a_{n}}{\sqrt{1-a_{n}}},\\ B&=a_{1}(1-a_{1})+a_{2}(1-a_{2})+\cdots+a_{n}(1-a_{n}). \end{aligned}\]By Hölder’s inequality we have
\[A^{2}B \ge (a_{1}+a_{2}+\cdots+a_{n})^{3}=1. \qquad (1)\]Applying \(QM\ge AM\) we deduce
\[B=1-(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}) \le 1-\frac{(a_{1}+a_{2}+\cdots+a_{n})^{2}}{n} =\frac{n-1}{n}. \qquad (2)\]By (1) and (2) we obtain
\[\frac{n-1}{n}\cdot A^{2} \ge A^{2}B \ge 1, \quad\text{i.e.}\quad A \ge \sqrt{\frac{n}{n-1}}.\]Equality 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.
Therefore, 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 \((x+y+z)^{2}\ge 3(xy+yz+zx)\) for
\[x=a^{2}+bc-ab,\quad y=b^{2}+ca-bc,\quad z=c^{2}+ab-ca,\]we obtain the required inequality.
Equality 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 \(C=3\).
Therefore, 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
\[(a+b)^{2}(1+2c)\left(2+3\frac{c}{a}\right)\left(2+3\frac{c}{b}\right)\ge 54c.\]By the Cauchy-Schwarz inequality and \(AM\ge GM\) we have
\[\begin{aligned} \left(2+3\frac{c}{a}\right)\left(2+3\frac{c}{b}\right) &\ge \left(2+\frac{3c}{\sqrt{ab}}\right)^{2} \ge \left(2+\frac{6c}{a+b}\right)^{2}\\ &=\frac{(2(a+b)+6c)^{2}}{(a+b)^{2}} =\frac{4(a+b+3c)^{2}}{(a+b)^{2}}. \end{aligned}\]Then we have
\[\begin{aligned} (a+b)^{2}(1+2c)\left(2+3\frac{c}{a}\right)\left(2+3\frac{c}{b}\right) &\ge (a+b)^{2}(1+2c)\cdot \frac{4(a+b+3c)^{2}}{(a+b)^{2}}\\ &=4(1+2c)(a+b+3c)^{2}. \end{aligned}\]If \(a+b+c=1\), then \(a+b=1-c\) and hence \(a+b+3c=1+2c\). Therefore,
\[4(1+2c)(a+b+3c)^{2}=4(1+2c)^{3},\]and it remains to prove that
\[4(1+2c)^{3}\ge 54c, \quad\text{i.e.}\quad (1+2c)^{3}\ge \frac{27c}{2}.\]By \(AM\ge GM\) we have
\[(1+2c)^{3}=\left(\frac{1}{2}+\frac{1}{2}+2c\right)^{3} \ge 27\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot 2c =\frac{27c}{2},\]as required.
Equality occurs iff \(a=b=\frac{3}{8}\) and \(c=\frac{1}{4}\). This gives the maximum value of \(C\) for which the inequality always holds, so \(C=54\) is maximal.
Therefore, 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
\[\begin{aligned} &f(a+b+c,\,ab+bc+ca,\,abc) \\ &= 9\left((a+b)(b+c)(c+a)\right)^2 \\ &\quad-4(ab+bc+ca)\Big((a+b)^2(b+c)^2+(b+c)^2(c+a)^2+(c+a)^2(a+b)^2\Big) \\ &= k(abc)^2+m\cdot abc+n, \end{aligned}\]where \(k\ge 0\) and \(k,m,n\) are quantities depending only on constants and on \(a+b+c\) and \(ab+bc+ca\) (which we treat as constants). Thus, the left-hand side is a sixth-degree symmetric polynomial in \(a,b,c\) and can be viewed as a quadratic polynomial in \(abc\) with nonnegative leading coefficient.
Let us explain this.
The expression
\[(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc\]has the form \(M-abc\), hence
\[9\left((a+b)(b+c)(c+a)\right)^2 =9(abc)^2-18M\cdot abc+9M^2,\]which is of the form \(k(abc)^2+m\cdot abc+n\).
Furthermore,
\[4(ab+bc+ca)\Big((a+b)^2(b+c)^2+(b+c)^2(c+a)^2+(c+a)^2(a+b)^2\Big)\]is a symmetric sixth-degree polynomial, and after fixing \(a+b+c\) and \(ab+bc+ca\) it can also be expressed in the form \(\tilde k(abc)^2+\tilde m\cdot abc+\tilde n\). Therefore the whole expression
\[f(a+b+c,\,ab+bc+ca,\,abc)\]has the form \(k(abc)^2+m\cdot abc+n\).
Hence the function achieves its minimum value when \((a-b)(b-c)(c-a)=0\) or when \(abc=0\).
If \((a-b)(b-c)(c-a)=0\), then without loss of generality we may assume \(a=c\), and the given inequality is equivalent to
\[\begin{aligned} &\left(a^{2}+2ab\right)\left(\frac{1}{4a^{2}}+\frac{2}{(a+b)^{2}}\right)\ge \frac{9}{4} \\ &\Leftrightarrow (a-b)^{2}\left(\frac{2a+b}{2a(a+b)^{2}}-\frac{1}{(a+b)^{2}}\right)\ge 0 \\ &\Leftrightarrow b(a-b)^{2}\ge 0, \end{aligned}\]as required.
If \(abc=0\), we may assume \(c=0\) and the given inequality becomes
\[\begin{aligned} &ab\left(\frac{1}{(a+b)^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)\ge \frac{9}{4} \\ &\Leftrightarrow (a-b)^{2}\left(\frac{1}{ab}-\frac{1}{4(a+b)^{2}}\right)\ge 0 \\ &\Leftrightarrow (a-b)^{2}\left(4a^{2}+4b^{2}+7ab\right)\ge 0, \end{aligned}\]and the problem is solved.
Equality 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.
Therefore, 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 \(AM \ge GM\) we have
\[\sqrt[n]{a_{1}a_{2}\cdots a_{n}} \le \frac{a_{1}+a_{2}+\cdots+a_{n}}{n} =\frac{1}{n},\]or
\[n \le \sqrt[n]{\frac{1}{a_{1}}\cdot\frac{1}{a_{2}}\cdots\frac{1}{a_{n}}}.\]Hence
\[n^{k} \le \sqrt[n]{a_{1}^{-k}a_{2}^{-k}\cdots a_{n}^{-k}} \le \frac{a_{1}^{-k}+a_{2}^{-k}+\cdots+a_{n}^{-k}}{n},\]i.e.
\[a_{1}^{-k}+a_{2}^{-k}+\cdots+a_{n}^{-k} \ge n^{k+1},\]as required.
Equality 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.
Therefore, 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, taking into account that
\[abc\le \frac{1}{27},\]we have
\[d\left(\frac{1}{27}-abc\right) > a^{3}+b^{3}+c^{3}-\frac{1}{9}.\]We may assume that
\[abc<\frac{1}{27}.\]Now we will reach a contradiction by proving that
\[a^{3}+b^{3}+c^{3}+abcd \ge \frac{1}{4}.\]It is sufficient to prove that
\[\frac{a^{3}+b^{3}+c^{3}-\frac{1}{9}}{\frac{1}{27}-abc}\cdot abc+a^{3}+b^{3}+c^{3}\ge \frac{1}{4}.\]But this inequality is equivalent to
\[4\sum a^{3}+15abc \ge 1.\]We now use the identity
\[\sum a^{3}=3abc+1-3\sum ab\]and reduce the problem to proving that
\[\sum ab \le \frac{1+9abc}{4},\]which is Schur’s inequality.
Equality holds when \(a=b=c=\frac{1}{3}\) (with \(d\) arbitrary), since then
\[a^{3}+b^{3}+c^{3}+abcd =3\left(\frac{1}{3}\right)^{3}+\left(\frac{1}{27}\right)d =\frac{1}{9}+\frac{d}{27}.\]Therefore, 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. $C = \frac{1}{32}$ (The original solution wrongly suggested $\frac{8}{729}$ as an answer.)
Let \(p=x+y+z=1\), \(q=xy+yz+zx\), \(r=xyz\). Then we have
\[\begin{aligned} x^{2}+y^{2} &=(x+y)^{2}-2xy \\ &=(1-z)^{2}-2xy \\ &=1-2z+z^{2}-2xy \\ &=1-z-z(1-z)-2xy \\ &=1-z-z(x+y)-2xy \\ &=1-z-q-xy. \end{aligned}\]Analogously we deduce
\[y^{2}+z^{2}=1-x-q-yz \quad\text{and}\quad z^{2}+x^{2}=1-y-q-zx.\]So the given inequality becomes
\[(1-z-q-xy)(1-x-q-yz)(1-y-q-zx)\le \frac{1}{32}. \qquad (1)\]After algebraic transformations we find that inequality (1) is equivalent to
\[q^{2}-2q^{3}-r(2+r-4q)\le \frac{1}{32}. \qquad (2)\]Assume that \(q\le \frac{1}{4}\). Using \(p^{3}-4pq+9r\ge 0\), it follows that
\[9r\ge 4q-1,\quad\text{i.e.}\quad r\ge \frac{4q-1}{9},\]and clearly \(q\le \frac{1}{3}\). It follows that
\[2+r-4q \ge 2+\frac{4q-1}{9}-4q =\frac{17-32q}{9} \ge \frac{17-\frac{32}{3}}{9} >0.\]So we have
\[\begin{aligned} q^{2}-2q^{3}-r(2+r-4q) &\le q^{2}-2q^{3} = q^{2}(1-2q) \\ &=\frac{q}{2}\cdot 2q(1-2q) \le \frac{q}{2}\left(\frac{2q+(1-2q)}{2}\right)^{2} =\frac{q}{8} \le \frac{1}{32}, \end{aligned}\]i.e. inequality (2) holds for \(q\le \frac{1}{4}\).
We need just to consider the case when \(q>\frac{1}{4}\). Let
\[f(r)=q^{2}-2q^{3}-r(2+r-4q). \qquad (3)\]Clearly \(r\ge \frac{4q-1}{9}\). Using \(pq-9r\ge 0\) it follows that \(9r\le q\), i.e. \(r\le \frac{q}{9}\). We have
\[f'(r)=4q-2-2r \le \frac{4}{3}-2-2r \le 0.\]This means that \(f\) is a strictly decreasing function on
\[\left(\frac{4q-1}{9},\,\frac{q}{9}\right),\]from which it follows that
\[f(r)\le f\left(\frac{4q-1}{9}\right) = q^{2}-2q^{3}-\frac{1}{81}(4q-1)(17-32q),\]i.e.
\[f(r)\le \frac{81\left(q^{2}-2q^{3}\right)-(4q-1)(17-32q)}{81}. \qquad (4)\]Let
\[g(q)=81\left(q^{2}-2q^{3}\right)-(4q-1)(17-32q). \qquad (5)\]Then
\[g'(q)=-486q^{2}+418q-100.\]Since \(\frac{1}{4}<q\le \frac{1}{3}\), we get
\[g'(q)=-486q^{2}+418q-100 <\frac{-486}{16}+\frac{418}{3}-100 <0.\]So \(g\) decreases on \(\left(\frac{1}{4},\frac{1}{3}\right)\), i.e.
\[g(q)<g\left(\frac{1}{4}\right)=\frac{81}{32}. \qquad (6)\]Finally by (3), (4), (5) and (6) we obtain
\[\begin{aligned} q^{2}-2q^{3}-r(2+r-4q) &= f(r) \le f\left(\frac{4q-1}{9}\right) \\ &=\frac{81\left(q^{2}-2q^{3}\right)-(4q-1)(17-32q)}{81} \\ &=\frac{g(q)}{81} <\frac{g\left(\frac{1}{4}\right)}{81} =\frac{\frac{81}{32}}{81} =\frac{1}{32}, \end{aligned}\]as required.
Equality in \(\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)\le \frac{1}{32}\) holds for permutations of \(\left(\frac12,\frac12,0\right)\), since
\[\left(\left(\frac12\right)^2+\left(\frac12\right)^2\right)\left(\left(\frac12\right)^2+0^2\right)\left(0^2+\left(\frac12\right)^2\right) =\left(\frac12\right)\left(\frac14\right)\left(\frac14\right)=\frac{1}{32}.\]If we additionally require \(x,y,z>0\), then equality is not attained but the bound \(\frac{1}{32}\) is sharp and is approached as one variable tends to \(0\).
Therefore, 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
\[\begin{aligned} 1 &=a^{3}+b^{3}+c^{3}-3abc \\ &=(a+b+c)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right) \\ &=\frac{a+b+c}{2}\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right). \end{aligned}\]Since
\[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\ge 0,\]we must have \(a+b+c>0\).
According to
\[(a+b+c)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)=1\]we deduce
\[(a+b+c)\left(a^{2}+b^{2}+c^{2}-\frac{(a+b+c)^{2}-a^{2}-b^{2}-c^{2}}{2}\right)=1,\]and hence
\[(a+b+c)\left(\frac{3}{2}(a^{2}+b^{2}+c^{2})-\frac{(a+b+c)^{2}}{2}\right)=1.\]Therefore,
\[3(a^{2}+b^{2}+c^{2})-(a+b+c)^{2}=\frac{2}{a+b+c},\]so
\[a^{2}+b^{2}+c^{2}=\frac{1}{3}\left((a+b+c)^{2}+\frac{2}{a+b+c}\right).\]Since \(a+b+c>0\) we may use \(AM \ge GM\) as follows:
\[(a+b+c)^{2}+\frac{1}{a+b+c}+\frac{1}{a+b+c} \ge 3\sqrt[3]{(a+b+c)^{2}\cdot\frac{1}{a+b+c}\cdot\frac{1}{a+b+c}} =3.\]Thus,
\[a^{2}+b^{2}+c^{2} =\frac{1}{3}\left((a+b+c)^{2}+\frac{2}{a+b+c}\right) \ge \frac{1}{3}\cdot 3 =1,\]as required.
Equality occurs iff
\[(a+b+c)^{2}=\frac{1}{a+b+c},\]i.e. iff \(a+b+c=1\). For example, \(a=b=c=\frac{1}{3}\) satisfies this, and then
\[a^{2}+b^{2}+c^{2}=3\left(\frac{1}{3}\right)^{2}=\frac{1}{3}.\]So the minimum value is \(\frac{1}{3}\), and the sharp constant is
\[C=\frac{1}{3}.\]Therefore, the answer is \(C=\frac{1}{3}\).
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 \(AM \ge GM\) we get
\[abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge 4\sqrt[4]{abc\cdot\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}} =4.\]However, equality in this step would require
\[abc=\frac{1}{a}=\frac{1}{b}=\frac{1}{c},\]which implies \(a=b=c=1\), contradicting \(a+b+c=1\).
So we need a sharper estimate.
Assume \(a,b,c>0\) and \(a+b+c=1\).
We aim to prove
Choose the constant \(81\) so that at the symmetric point \(a=b=c=\frac{1}{3}\) we have
\[abc=\frac{1}{81a}=\frac{1}{81b}=\frac{1}{81c}=\frac{1}{27}.\]Now rewrite the expression as
\[abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \left(abc+\frac{1}{81a}+\frac{1}{81b}+\frac{1}{81c}\right) +\frac{80}{81}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). \qquad (1)\]By \(AM \ge GM\),
\[abc+\frac{1}{81a}+\frac{1}{81b}+\frac{1}{81c} \ge 4\sqrt[4]{abc\cdot\frac{1}{81a}\cdot\frac{1}{81b}\cdot\frac{1}{81c}} =4\sqrt[4]{\frac{1}{81^{3}}} =\frac{4}{27}. \qquad (2)\]By \(AM \ge HM\),
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9}{a+b+c} =9. \qquad (3)\]Combining \((1),(2),(3)\) gives
\[abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{4}{27}+\frac{80}{81}\cdot 9 =\frac{4}{27}+\frac{80}{9} =\frac{244}{27}.\]Equality holds when \(a=b=c=\frac{1}{3}\), which satisfies \(a+b+c=1\) and yields
\[abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =\frac{1}{27}+9 =\frac{244}{27}.\]Therefore, the sharp constant 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 \(a=2,\ b=3,\ c=4\).
Using \(AM \ge GM\) we get
\[a+\frac{4}{a}\ge 2\sqrt{a\cdot \frac{4}{a}}=4,\quad b+\frac{9}{b}\ge 2\sqrt{b\cdot \frac{9}{b}}=6,\quad c+\frac{16}{c}\ge 2\sqrt{c\cdot \frac{16}{c}}=8.\]i.e.
\[\frac{3}{4}\left(a+\frac{4}{a}\right)\ge 3,\quad \frac{1}{2}\left(b+\frac{9}{b}\right)\ge 3,\quad \frac{1}{4}\left(c+\frac{16}{c}\right)\ge 2.\]Adding the last three inequalities we have
\[\frac{3}{4}a+\frac{1}{2}b+\frac{1}{4}c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}\ge 8. \qquad (1)\]Using \(a+2b+3c\ge 20\) we obtain
\[\frac{1}{4}a+\frac{1}{2}b+\frac{3}{4}c\ge 5. \qquad (2)\]Finally, after adding (1) and (2) we get
\[a+b+c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}\ge 13,\]as desired.
Equality holds when \(a=2\), \(b=3\), \(c=4\), since at this point
\[a+2b+3c=2+6+12=20\]and
\[S=2+3+4+\frac{3}{2}+\frac{9}{2\cdot 3}+\frac{4}{4} =9+\frac{3}{2}+\frac{3}{2}+1 =13.\]Thus the minimum value of \(S\) is \(13\), and the sharp constant is \(C=13\).
Therefore, 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
\[p_{2}=\frac{u}{\binom{4}{2}}=\frac{u}{6} \quad\text{and}\quad p_{3}=\frac{v}{\binom{4}{3}}=\frac{v}{4}.\]By Maclaurin’s inequality we have
\[p_{2}^{\frac{1}{2}} \ge p_{3}^{\frac{1}{3}} \quad\Leftrightarrow\quad p_{2}^{3}\ge p_{3}^{2} \quad\Leftrightarrow\quad \left(\frac{u}{6}\right)^{3}\ge \left(\frac{v}{4}\right)^{2} \quad\Leftrightarrow\quad 2u^{3}\ge 27v^{2}.\]Equality holds when \(a=b=c=d\), i.e. all variables are equal, in which case Maclaurin’s inequality becomes an equality. Hence the sharp constant is
\[C=27.\]Therefore, 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
\[\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y} =\frac{1}{2}\left(\frac{xy}{z}+\frac{yz}{x}\right) +\frac{1}{2}\left(\frac{yz}{x}+\frac{zx}{y}\right) +\frac{1}{2}\left(\frac{zx}{y}+\frac{xy}{z}\right). \qquad (1)\]Since \(AM \ge GM\) we have
\[\frac{1}{2}\left(\frac{xy}{z}+\frac{yz}{x}\right) \ge \sqrt{\frac{xy}{z}\cdot\frac{yz}{x}} =\sqrt{y^{2}} =y.\]Analogously we get
\[\frac{1}{2}\left(\frac{yz}{x}+\frac{zx}{y}\right)\ge z \quad\text{and}\quad \frac{1}{2}\left(\frac{zx}{y}+\frac{xy}{z}\right)\ge x.\]Adding these three inequalities we obtain
\[\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\ge x+y+z=1.\]Equality holds if and only if
\[\frac{xy}{z}=\frac{yz}{x}=\frac{zx}{y},\]which is equivalent to \(x=y=z\). Under \(x+y+z=1\) this gives
\[x=y=z=\frac{1}{3}.\]Thus the minimum value of the expression is \(1\), and the sharp constant is \(C=1\).
Therefore, 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 \(a\ge b\ge c\ge d\).
Let us denote
\[\begin{aligned} f(a,b,c,d) &=a^{4}+b^{4}+c^{4}+d^{4}+2abcd-a^{2}b^{2}-b^{2}c^{2}-c^{2}d^{2}-d^{2}a^{2}-a^{2}c^{2}-b^{2}d^{2} \\ &=a^{4}+b^{4}+c^{4}+d^{4}+2abcd-a^{2}c^{2}-b^{2}d^{2}-\left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right). \end{aligned}\]We have
\[\begin{aligned} &f(a,b,c,d)-f(\sqrt{ac},\,b,\,\sqrt{ac},\,d) \\ &=\Big(a^{4}+b^{4}+c^{4}+d^{4}+2abcd-a^{2}c^{2}-b^{2}d^{2}-\left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right)\Big) \\ &\quad-\Big((ac)^{2}+b^{4}+(ac)^{2}+d^{4}+2abcd-(ac)^{2}-b^{2}d^{2}-2ac\left(b^{2}+d^{2}\right)\Big) \\ &=a^{4}+c^{4}-2a^{2}c^{2}-\left(b^{2}+d^{2}\right)\left(a^{2}+c^{2}-2ac\right) \\ &=(a^{2}-c^{2})^{2}-\left(b^{2}+d^{2}\right)(a-c)^{2} \\ &=(a-c)^{2}\left((a+c)^{2}-\left(b^{2}+d^{2}\right)\right)\ge 0, \end{aligned}\]because \(a\ge b\ge c\ge d\ge 0\) implies \(a+c\ge b\ge 0\) and hence \((a+c)^{2}\ge b^{2}\ge b^{2}+d^{2}\).
Thus
\[f(a,b,c,d)\ge f(\sqrt{ac},\,b,\,\sqrt{ac},\,d).\]By the \(uvw\)-type (smoothing) principle, it is enough to prove \(f(a,b,c,d)\ge 0\) in the case when \(a=b=c=t\ge d\).
We have
\[\begin{aligned} f(t,t,t,d)\ge 0 &\Leftrightarrow 3t^{4}+d^{4}+2t^{3}d \ge 3t^{4}+3t^{2}d^{2} \\ &\Leftrightarrow d^{4}+2t^{3}d \ge 3t^{2}d^{2}. \end{aligned}\]If \(d=0\) the inequality is obvious. For \(d>0\), divide by \(d^{2}\) to get
\[d^{2}+2t^{3}\cdot\frac{1}{d}\ge 3t^{2}.\]Now apply \(AM\ge GM\) to the three positive numbers \(d^{2}\), \(t^{3}/d\), \(t^{3}/d\):
\[d^{2}+\frac{t^{3}}{d}+\frac{t^{3}}{d}\ge 3\sqrt[3]{d^{2}\cdot\frac{t^{3}}{d}\cdot\frac{t^{3}}{d}}=3t^{2},\]which proves \(f(t,t,t,d)\ge 0\).
Equality occurs when \(a=b=c=d\) or when \(d=0\) and \(a=b=c\) (up to permutation).
Therefore, the sharp constant 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
\[\frac{LP}{R^{3}} =\frac{(a+b+c)abc}{4R^{4}}.\]Using \(a=2R\sin\alpha\), \(b=2R\sin\beta\), \(c=2R\sin\gamma\), we get
\[a+b+c=2R(\sin\alpha+\sin\beta+\sin\gamma)\]and
\[abc=8R^{3}\sin\alpha\sin\beta\sin\gamma.\]Hence
\[\frac{LP}{R^{3}} =\frac{2R(\sin\alpha+\sin\beta+\sin\gamma)\cdot 8R^{3}\sin\alpha\sin\beta\sin\gamma}{4R^{4}} =4(\sin\alpha+\sin\beta+\sin\gamma)\sin\alpha\sin\beta\sin\gamma. \qquad (1)\]By \(AM \ge GM\) we have
\[\sin\alpha\sin\beta\sin\gamma \le \left(\frac{\sin\alpha+\sin\beta+\sin\gamma}{3}\right)^{3}.\]So by (1) we get
\[\frac{LP}{R^{3}} \le \frac{4(\sin\alpha+\sin\beta+\sin\gamma)^{4}}{27}. \qquad (2)\]The function \(f(x)=\sin x\) is concave on \([0,\pi]\), so by Jensen’s inequality we have
\[\frac{\sin\alpha+\sin\beta+\sin\gamma}{3} \le \sin\left(\frac{\alpha+\beta+\gamma}{3}\right) =\sin\left(\frac{\pi}{3}\right) =\frac{\sqrt{3}}{2}.\]Thus
\[\sin\alpha+\sin\beta+\sin\gamma \le \frac{3\sqrt{3}}{2}.\]Finally from (2) we obtain
\[\frac{LP}{R^{3}} \le \frac{4}{27}\left(\frac{3\sqrt{3}}{2}\right)^{4} =\frac{4}{27}\cdot\frac{729}{16} =\frac{27}{4}.\]Equality holds when \(\alpha=\beta=\gamma=\frac{\pi}{3}\), i.e. \(a=b=c\) and the triangle is equilateral. Therefore the sharp constant 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
\[a=\frac{x}{y},\quad b=\frac{t}{x},\quad c=\frac{z}{t},\quad d=\frac{y}{z},\]the given inequality becomes
\[\frac{x}{z+t}+\frac{y}{x+t}+\frac{z}{x+y}+\frac{t}{z+y}\ge 2.\]By the Cauchy-Schwarz inequality we have
\[\begin{aligned} \frac{x}{z+t}+\frac{y}{x+t}+\frac{z}{x+y}+\frac{t}{z+y} &=\frac{x^{2}}{xz+xt}+\frac{y^{2}}{yx+yt}+\frac{z^{2}}{zx+zy}+\frac{t^{2}}{tz+ty} \\ &\ge \frac{(x+y+z+t)^{2}}{(xz+xt)+(yx+yt)+(zx+zy)+(tz+ty)} \\ &=\frac{(x+y+z+t)^{2}}{2xz+2yt+xt+xy+zy+tz}. \end{aligned}\]Hence it suffices to prove that
\[\frac{(x+y+z+t)^{2}}{2xz+2yt+xt+xy+zy+tz}\ge 2,\]which is equivalent to
\[(x+y+z+t)^{2}\ge 4xz+4yt+2xt+2xy+2zy+2tz.\]Expanding and simplifying gives
\[(x-z)^{2}+(y-t)^{2}\ge 0,\]which is true.
Equality occurs when \(x=z\) and \(y=t\), which corresponds to
\[a=\frac{x}{y}=\frac{z}{t}=c \quad\text{and}\quad b=\frac{t}{x}=\frac{y}{z}=d.\]Thus the minimum value of the left-hand side is \(2\), so the sharp constant is \(C=2\).
Therefore, 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
\[\begin{aligned} x^{4}+y^{4}+z^{4}-4xyz+1 &=\left(x^{4}-2x^{2}+1\right)+\left(y^{4}-2y^{2}z^{2}+z^{4}\right)+\left(2y^{2}z^{2}-4xyz+2x^{2}\right) \\ &=(x^{2}-1)^{2}+(y^{2}-z^{2})^{2}+2(yz-x)^{2} \\ &\ge 0. \end{aligned}\]So it follows that
\[x^{4}+y^{4}+z^{4}\ge 4xyz-1.\]Equality holds when
\[x^{2}=1,\quad y^{2}=z^{2},\quad yz=x,\]which gives \(x=1\) and \(y=z=1\) or \(x=-1\) and \(y=z=-1\). In particular, taking \(x=y=z=1\) yields equality:
\[1+1+1=4\cdot 1\cdot 1\cdot 1-1.\]Thus the sharp constant is
\[C=-1.\]Therefore, 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 \(a,b,c>0\) and \(a+b+c=1\) it follows that \(0<a,b,c<1\).
The given inequality is symmetric, so without loss of generality we may assume that \(a\le b\le c\).
Then we have
\[1+a^{2}\le 1+b^{2}\le 1+c^{2} \quad\text{and}\quad 1-c^{2}\le 1-b^{2}\le 1-a^{2},\]hence
\[\frac{1}{1-a^{2}}\le \frac{1}{1-b^{2}}\le \frac{1}{1-c^{2}}.\]Now by Chebyshev’s inequality we have
\[\begin{aligned} A &=\frac{1+a^{2}}{1-a^{2}}+\frac{1+b^{2}}{1-b^{2}}+\frac{1+c^{2}}{1-c^{2}} \\ &\ge \frac{1}{3}\left((1+a^{2})+(1+b^{2})+(1+c^{2})\right) \left(\frac{1}{1-a^{2}}+\frac{1}{1-b^{2}}+\frac{1}{1-c^{2}}\right), \end{aligned}\]i.e.
\[A\ge \frac{a^{2}+b^{2}+c^{2}+3}{3} \left(\frac{1}{1-a^{2}}+\frac{1}{1-b^{2}}+\frac{1}{1-c^{2}}\right). \qquad (1)\]Also we have the well-known inequality
\[a^{2}+b^{2}+c^{2}\ge \frac{(a+b+c)^{2}}{3}=\frac{1}{3}.\]Therefore by (1) we obtain
\[A\ge \frac{\frac{1}{3}+3}{3} \left(\frac{1}{1-a^{2}}+\frac{1}{1-b^{2}}+\frac{1}{1-c^{2}}\right) =\frac{10}{9}\left(\frac{1}{1-a^{2}}+\frac{1}{1-b^{2}}+\frac{1}{1-c^{2}}\right). \qquad (2)\]Since \(1-a^{2},1-b^{2},1-c^{2}>0\), by using \(AM \ge HM\) we deduce
\[\frac{1}{1-a^{2}}+\frac{1}{1-b^{2}}+\frac{1}{1-c^{2}} \ge \frac{9}{(1-a^{2})+(1-b^{2})+(1-c^{2})} =\frac{9}{3-(a^{2}+b^{2}+c^{2})}.\]Using \(a^{2}+b^{2}+c^{2}\ge \frac{1}{3}\) gives
\[\frac{1}{1-a^{2}}+\frac{1}{1-b^{2}}+\frac{1}{1-c^{2}} \ge \frac{9}{3-\frac{1}{3}}=\frac{27}{8}. \qquad (3)\]Finally from (2) and (3) we get
\[A\ge \frac{10}{9}\cdot\frac{27}{8}=\frac{15}{4}.\]Equality holds when \(a=b=c=\frac{1}{3}\). Indeed, then
\[A=3\cdot\frac{1+\left(\frac{1}{3}\right)^{2}}{1-\left(\frac{1}{3}\right)^{2}} =3\cdot\frac{\frac{10}{9}}{\frac{8}{9}} =3\cdot\frac{5}{4} =\frac{15}{4}.\]Therefore, the sharp constant 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
\[\begin{aligned} (1+a)(a+b)(b+c)(c+16) &=\left(1+\frac{a}{2}+\frac{a}{2}\right) \left(a+\frac{b}{2}+\frac{b}{2}\right) \left(b+\frac{c}{2}+\frac{c}{2}\right) \left(c+8+8\right) \\ &\ge \left(3\sqrt[3]{1\cdot\frac{a}{2}\cdot\frac{a}{2}}\right) \left(3\sqrt[3]{a\cdot\frac{b}{2}\cdot\frac{b}{2}}\right) \left(3\sqrt[3]{b\cdot\frac{c}{2}\cdot\frac{c}{2}}\right) \left(3\sqrt[3]{c\cdot 8\cdot 8}\right) \\ &= 3\sqrt[3]{\frac{a^{2}}{4}}\cdot 3\sqrt[3]{\frac{ab^{2}}{4}} \cdot 3\sqrt[3]{\frac{bc^{2}}{4}}\cdot 3\sqrt[3]{64c} \\ &=81\sqrt[3]{\frac{a^{3}b^{3}c^{3}}{64}\cdot 64} =81\sqrt[3]{a^{3}b^{3}c^{3}} =81abc. \end{aligned}\]Thus
\[\frac{abc}{(1+a)(a+b)(b+c)(c+16)}\le \frac{1}{81}.\]Equality holds when
\[1=\frac{a}{2},\quad a=\frac{b}{2},\quad b=\frac{c}{2},\quad c=8,\]which gives
\[a=2,\quad b=4,\quad c=8.\]Indeed, then
\[\frac{abc}{(1+a)(a+b)(b+c)(c+16)} =\frac{2\cdot 4\cdot 8}{(1+2)(2+4)(4+8)(8+16)} =\frac{64}{3\cdot 6\cdot 12\cdot 24} =\frac{1}{81}.\]Therefore, the sharp constant 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
\[4b\sqrt{c}-c\sqrt{a}>4\sqrt{c}-2\sqrt{c}=2\sqrt{c}>0.\]Analogously we get \(4c\sqrt{a}-a\sqrt{b}>0\) and \(4a\sqrt{b}-b\sqrt{c}>0\).
We’ll prove that
\[\frac{b\sqrt{a}}{4b\sqrt{c}-c\sqrt{a}}\ge \frac{a}{a+b+c}. \qquad (1)\]Since \(4b\sqrt{c}-c\sqrt{a}>0\), inequality (1) is equivalent to
\[b(a+b+c)\ge a(4b\sqrt{c}-c\sqrt{a})\cdot\frac{\sqrt{a}}{a},\]i.e.
\[b(a+b+c)\ge 4b\sqrt{ac}-ac.\]This can be rewritten as
\[ab+b^{2}+bc+ac \ge 4b\sqrt{ac},\]or
\[(a+b)(b+c)\ge 4b\sqrt{ac},\]which is true by \(AM\ge GM\):
\[a+b\ge 2\sqrt{ab},\quad b+c\ge 2\sqrt{bc} \quad\Rightarrow\quad (a+b)(b+c)\ge 4b\sqrt{ac}.\]Similarly we deduce that
\[\frac{c\sqrt{b}}{4c\sqrt{a}-a\sqrt{b}}\ge \frac{b}{a+b+c}, \qquad (2)\]and
\[\frac{a\sqrt{c}}{4a\sqrt{b}-b\sqrt{c}}\ge \frac{c}{a+b+c}. \qquad (3)\]Adding (1), (2) and (3) we get
\[\frac{b\sqrt{a}}{4b\sqrt{c}-c\sqrt{a}} +\frac{c\sqrt{b}}{4c\sqrt{a}-a\sqrt{b}} +\frac{a\sqrt{c}}{4a\sqrt{b}-b\sqrt{c}} \ge \frac{a+b+c}{a+b+c}=1.\]Equality requires equality in each \(AM\ge GM\) step, hence
\[a=b=c.\]For \(a=b=c=t\in(1,2)\), each denominator is
\[4t\sqrt{t}-t\sqrt{t}=3t\sqrt{t},\]so each term equals
\[\frac{t\sqrt{t}}{3t\sqrt{t}}=\frac{1}{3},\]and the sum is \(1\).
Therefore, the sharp constant 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 \(a\ge b\ge c\). Let
\[x=\frac{1}{a},\quad y=\frac{1}{b},\quad z=\frac{1}{c}.\]Then clearly
\[xyz=1.\]We have
\[\begin{aligned} \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} &=\frac{x^{3}}{\frac{1}{y}+\frac{1}{z}}+\frac{y^{3}}{\frac{1}{z}+\frac{1}{x}}+\frac{z^{3}}{\frac{1}{x}+\frac{1}{y}} \\ &=\frac{x^{3}}{\frac{y+z}{yz}}+\frac{y^{3}}{\frac{z+x}{zx}}+\frac{z^{3}}{\frac{x+y}{xy}} \\ &=\frac{x^{3}yz}{y+z}+\frac{y^{3}zx}{z+x}+\frac{z^{3}xy}{x+y} \\ &=\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}. \end{aligned}\]Since \(c\le b\le a\) we have \(x\le y\le z\). So
\[x+y\le x+z\le y+z \quad\Rightarrow\quad \frac{1}{y+z}\le \frac{1}{z+x}\le \frac{1}{x+y},\]and also
\[x^{2}\le y^{2}\le z^{2}.\]Thus, by the rearrangement inequality,
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \ge \frac{x^{2}}{x+y}+\frac{y^{2}}{z+x}+\frac{z^{2}}{y+z}.\]Now apply \(AM\ge GM\) termwise:
\[\frac{x^{2}}{x+y}\ge \frac{x^{2}}{2x}=\frac{x}{2},\quad \frac{y^{2}}{x+z}\ge \frac{y^{2}}{2y}=\frac{y}{2},\quad \frac{z^{2}}{y+z}\ge \frac{z^{2}}{2z}=\frac{z}{2}.\]Adding these gives
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \ge \frac{x+y+z}{2}.\]Therefore,
\[2\left(\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) =2\left(\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}\right) \ge x+y+z.\]Finally, by \(AM\ge GM\),
\[x+y+z\ge 3\sqrt[3]{xyz}=3,\]so
\[\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\ge \frac{3}{2},\]as required.
Equality holds when \(x=y=z=1\), i.e. \(a=b=c=1\). In this case,
\[\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} =3\cdot\frac{1}{1^{3}(1+1)}=\frac{3}{2},\]so the bound is sharp and the sharp constant is
\[C=\frac{3}{2}.\]Therefore, 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
\[\begin{aligned} 2x^{4}+1-2x^{3}-x^{2} &=1-x^{2}-2x^{3}(1-x) \\ &=(1-x)(1+x)-2x^{3}(1-x) \\ &=(1-x)\left(1+x-2x^{3}\right) \\ &=(1-x)\left(x(1-x^{2})+1-x^{3}\right) \\ &=(1-x)\left(x(1-x)(1+x)+(1-x)(1+x+x^{2})\right) \\ &=(1-x)^{2}\left(x(1+x)+1+x+x^{2}\right) \\ &=(1-x)^{2}\left((x+1)^{2}+x^{2}\right)\ge 0. \end{aligned}\]Equality occurs if and only if \(x=1\), since then \((1-x)^{2}=0\). Thus the sharp constant is
\[C=1.\]Therefore, 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 \(AM \ge GM\) we have
\[\frac{a^{3}+2}{b+2} =\frac{a^{3}+1+1}{b+2} \ge \frac{3\sqrt[3]{a^{3}\cdot 1\cdot 1}}{b+2} =\frac{3a}{b+2}.\]Similarly we get
\[\frac{b^{3}+2}{c+2}\ge \frac{3b}{c+2} \quad\text{and}\quad \frac{c^{3}+2}{a+2}\ge \frac{3c}{a+2}.\]Therefore
\[\frac{a^{3}+2}{b+2}+\frac{b^{3}+2}{c+2}+\frac{c^{3}+2}{a+2} \ge 3\left(\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}\right). \qquad (1)\]Applying the Cauchy-Schwarz inequality we obtain
\[\begin{aligned} \frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2} &=\frac{a^{2}}{a(b+2)}+\frac{b^{2}}{b(c+2)}+\frac{c^{2}}{c(a+2)} \\ &\ge \frac{(a+b+c)^{2}}{a(b+2)+b(c+2)+c(a+2)} \\ &=\frac{(a+b+c)^{2}}{ab+bc+ca+2(a+b+c)}. \end{aligned} \qquad (2)\]Since
\[(a+b+c)^{2}\ge 3(ab+bc+ca),\]we deduce that
\[ab+bc+ca\le \frac{(a+b+c)^{2}}{3}. \qquad (3)\]From (2) and (3) we get
\[\begin{aligned} \frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2} &\ge \frac{(a+b+c)^{2}}{ab+bc+ca+2(a+b+c)} \\ &\ge \frac{(a+b+c)^{2}}{\frac{(a+b+c)^{2}}{3}+2(a+b+c)} \\ &=\frac{3(a+b+c)^{2}}{(a+b+c)^{2}+6(a+b+c)} =\frac{3(a+b+c)}{(a+b+c)+6}. \end{aligned} \qquad (4)\]Finally by (1), (4), and since \(a+b+c=3\) we obtain
\[\frac{a^{3}+2}{b+2}+\frac{b^{3}+2}{c+2}+\frac{c^{3}+2}{a+2} \ge 3\left(\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}\right) \ge \frac{9(a+b+c)}{(a+b+c)+6} =\frac{27}{9} =3,\]as required.
Equality holds when \(a=b=c=1\), since then each term equals
\[\frac{1^{3}+2}{1+2}=1,\]so the sum is \(3\). Hence the minimum value is \(3\) and the sharp constant is
\[C=3.\]Therefore, 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:
\[l_{\alpha}=\frac{2\sqrt{bc}}{b+c}\sqrt{s(s-a)},\quad l_{\beta}=\frac{2\sqrt{ca}}{c+a}\sqrt{s(s-b)},\quad l_{\gamma}=\frac{2\sqrt{ab}}{a+b}\sqrt{s(s-c)}.\]From the obvious inequality
\[\frac{2\sqrt{xy}}{x+y}\le 1\]and the previous identities we obtain
\[l_{\alpha}\le \sqrt{s(s-a)},\quad l_{\beta}\le \sqrt{s(s-b)},\quad l_{\gamma}\le \sqrt{s(s-c)}. \qquad (1)\]Also,
\[h_{a}\le l_{\alpha},\quad h_{b}\le l_{\beta},\quad h_{c}\le l_{\gamma}. \qquad (2)\]So we have
\[\begin{aligned} \frac{l_{\alpha}}{a}+\frac{l_{\beta}}{b}+\frac{l_{\gamma}}{c} &=\frac{l_{\alpha}h_{a}}{2P}+\frac{l_{\beta}h_{b}}{2P}+\frac{l_{\gamma}h_{c}}{2P} \\ &\stackrel{(2)}{\le}\frac{l_{\alpha}^{2}+l_{\beta}^{2}+l_{\gamma}^{2}}{2P} \\ &\stackrel{(1)}{\le}\frac{s(s-a)+s(s-b)+s(s-c)}{2P} \\ &=\frac{s\left(3s-(a+b+c)\right)}{2P} =\frac{s(3s-2s)}{2P} =\frac{s^{2}}{2P}. \end{aligned}\]Since \(P=rs\), we get
\[\frac{s^{2}}{2P}=\frac{s^{2}}{2rs}=\frac{s}{2r}.\]Hence
\[\frac{l_{\alpha}}{a}+\frac{l_{\beta}}{b}+\frac{l_{\gamma}}{c}\le \frac{s}{2r}.\]Equality occurs if and only if the triangle is equilateral, since equality in (1) requires \(a=b=c\) and then (2) is also equality. Therefore the sharp constant 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 \(a=x+y\), \(b=y+z\), \(c=z+x\), where \(x,y,z>0\). Then
\[a^{2}+b^{2}+c^{2} =(x+y)^{2}+(y+z)^{2}+(z+x)^{2}.\]Also,
\[ab+bc+ca =(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y).\]We have
\[\begin{aligned} 2(ab+bc+ca)-(a^{2}+b^{2}+c^{2}) &=2\Big((x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)\Big) \\ &\quad-\Big((x+y)^{2}+(y+z)^{2}+(z+x)^{2}\Big) \\ &=2(xy+yz+zx). \end{aligned}\]Since \(x,y,z>0\), it follows that \(xy+yz+zx>0\), hence
\[a^{2}+b^{2}+c^{2}<2(ab+bc+ca).\]Thus the sharp constant is \(C=2\). Equality is not attained for \(x,y,z>0\), but it is approached when the triangle becomes degenerate, e.g. when one of \(x,y,z\) tends to \(0^{+}\).
Therefore, 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
\[\frac{\sqrt[3]{ab+bc}+\sqrt[3]{bc+ca}+\sqrt[3]{ca+ab}}{3} \le \sqrt[3]{\frac{(ab+bc)+(bc+ca)+(ca+ab)}{3}},\]i.e.
\[\sqrt[3]{ab+bc}+\sqrt[3]{bc+ca}+\sqrt[3]{ca+ab} \le 3\sqrt[3]{\frac{2(ab+bc+ca)}{3}} =\sqrt[3]{18(ab+bc+ca)}. \qquad (1)\]Since
\[ab+bc+ca\le \frac{(a+b+c)^{2}}{3}=\frac{6^{2}}{3}=12,\]by (1) we obtain
\[\sqrt[3]{ab+bc}+\sqrt[3]{bc+ca}+\sqrt[3]{ca+ab} \le \sqrt[3]{18\cdot 12} =\sqrt[3]{216} =6.\]Equality holds when \(a=b=c=2\), since then \(a+b+c=6\) and
\[\sqrt[3]{ab+bc}+\sqrt[3]{bc+ca}+\sqrt[3]{ca+ab} =3\sqrt[3]{4+4} =3\sqrt[3]{8} =6.\]Therefore, the sharp constant 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
\[\frac{1}{x_{1}+1998}+\frac{1}{x_{2}+1998}+\cdots+\frac{1}{x_{n}+1998} =\frac{1}{1998}\]becomes
\[a_{1}+a_{2}+\cdots+a_{n}=1.\]We need to show that
\[\left(\frac{1}{a_{1}}-1\right)\left(\frac{1}{a_{2}}-1\right)\cdots\left(\frac{1}{a_{n}}-1\right) \ge (n-1)^{n}. \qquad (1)\]We have
\[\frac{1}{a_{i}}-1=\frac{1-a_{i}}{a_{i}} =\frac{a_{1}+\cdots+a_{i-1}+a_{i+1}+\cdots+a_{n}}{a_{i}}.\]By \(AM\ge GM\) applied to the \(n-1\) numbers
\[a_{1},\ldots,a_{i-1},a_{i+1},\ldots,a_{n},\]we get
\[a_{1}+\cdots+a_{i-1}+a_{i+1}+\cdots+a_{n} \ge (n-1)\sqrt[n-1]{a_{1}\cdots a_{i-1}a_{i+1}\cdots a_{n}}.\]Hence
\[\frac{1}{a_{i}}-1 \ge (n-1)\sqrt[n-1]{\frac{a_{1}\cdots a_{i-1}a_{i+1}\cdots a_{n}}{a_{i}^{\,n-1}}}.\]Multiplying these inequalities for \(i=1,2,\ldots,n\), we obtain
\[\prod_{i=1}^{n}\left(\frac{1}{a_{i}}-1\right) \ge (n-1)^{n}\sqrt[n-1]{\prod_{i=1}^{n}\frac{a_{1}\cdots a_{i-1}a_{i+1}\cdots a_{n}}{a_{i}^{\,n-1}}}.\]But
\[\prod_{i=1}^{n}\left(a_{1}\cdots a_{i-1}a_{i+1}\cdots a_{n}\right) =(a_{1}a_{2}\cdots a_{n})^{n-1},\]so the radical becomes
\[\sqrt[n-1]{\frac{(a_{1}a_{2}\cdots a_{n})^{n-1}}{(a_{1}a_{2}\cdots a_{n})^{n-1}}}=1.\]Therefore,
\[\prod_{i=1}^{n}\left(\frac{1}{a_{i}}-1\right)\ge (n-1)^{n},\]which is exactly (1).
Equality holds when \(a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n}\). Then
\[\frac{1998}{x_{i}+1998}=\frac{1}{n} \quad\Rightarrow\quad x_{i}=1998(n-1)\]for all \(i\). Hence
\[\sqrt[n]{x_{1}x_{2}\cdots x_{n}}=1998(n-1),\]and this is the minimal value.
Therefore, the sharp constant 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. $C = 1$
It follows by summing the inequalities
\[\begin{aligned} &\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}\ge \frac{1}{1+ab},\\ &\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}}\ge \frac{1}{1+cd}. \end{aligned}\]The first of these inequalities follows from
\[\begin{aligned} \frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}-\frac{1}{1+ab} &=\frac{(1+b)^{2}(1+ab)+(1+a)^{2}(1+ab)-(1+a)^{2}(1+b)^{2}}{(1+a)^{2}(1+b)^{2}(1+ab)} \\ &=\frac{ab(a-b)^{2}+(ab-1)^{2}}{(1+a)^{2}(1+b)^{2}(1+ab)} \\ &\ge 0. \end{aligned}\]Similarly,
\[\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}}-\frac{1}{1+cd} =\frac{cd(c-d)^{2}+(cd-1)^{2}}{(1+c)^{2}(1+d)^{2}(1+cd)}\ge 0.\]Adding the two inequalities gives
\[\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \ge \frac{1}{1+ab}+\frac{1}{1+cd}.\]If in addition \(abcd=1\), then \(cd=\frac{1}{ab}\) and
\[\frac{1}{1+ab}+\frac{1}{1+cd} =\frac{1}{1+ab}+\frac{1}{1+\frac{1}{ab}} =\frac{1}{1+ab}+\frac{ab}{1+ab} =1.\]Hence
\[\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \ge 1.\]Equality holds when \(a=b=c=d=1\), which satisfies \(abcd=1\), and then each term equals \(\frac{1}{4}\) so the sum is \(1\). Therefore the sharp constant 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
\[A=\left(\frac{3-x}{1-x}\right)^{2}+\left(\frac{3-y}{1-y}\right)^{2}+\left(\frac{3-z}{1-z}\right)^{2}-7.\]We have \(\frac{3-x}{1-x}=\frac{(1-x)+2}{1-x}=1+\frac{2}{1-x},\)
so
\[A=\left(1+\frac{2}{1-x}\right)^{2}+\left(1+\frac{2}{1-y}\right)^{2}+\left(1+\frac{2}{1-z}\right)^{2}-7.\]Let
\[a=\frac{1}{1-x},\quad b=\frac{1}{1-y},\quad c=\frac{1}{1-z}.\]Then
\[A=(1+2a)^{2}+(1+2b)^{2}+(1+2c)^{2}-7,\]i.e.
\[A=4a^{2}+4b^{2}+4c^{2}+4a+4b+4c-4. \qquad (1)\]Furthermore, the condition \(xyz=1\) is equivalent to
\[(1-x)(1-y)(1-z)=1-x-y-z+xy+yz+zx-xyz,\]and using \(xyz=1\) this becomes
\[(1-x)(1-y)(1-z)=xy+yz+zx-(x+y+z).\]In terms of \(a,b,c\), since \(1-x=\frac{1}{a}\), etc., we have
\[\frac{1}{abc}= \left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right),\]which simplifies to
\[ab+bc+ca=a+b+c-1. \qquad (2)\]Using (1) and (2) we get
\[\begin{aligned} A &=4(a^{2}+b^{2}+c^{2})+4(a+b+c-1) \\ &=4(a^{2}+b^{2}+c^{2})+4(ab+bc+ca) \\ &=4(a^{2}+b^{2}+c^{2}+ab+bc+ca) \\ &=2\left((a+b)^{2}+(b+c)^{2}+(c+a)^{2}\right) \ge 0. \end{aligned}\]Equality holds if and only if
\[a+b=0,\quad b+c=0,\quad c+a=0,\]which implies \(a=b=c=0\), impossible because \(a=\frac{1}{1-x}\), \(b=\frac{1}{1-y}\), \(c=\frac{1}{1-z}\) are defined and nonzero.
Therefore,
\[A>0,\]so
\[\left(\frac{3-x}{1-x}\right)^{2}+\left(\frac{3-y}{1-y}\right)^{2}+\left(\frac{3-z}{1-z}\right)^{2}>7.\]Moreover, the bound \(7\) is sharp: taking \(x=y=t\) and \(z=\frac{1}{t^{2}}\) with \(t\to 0^{+}\) gives \(a\to 1\), \(b\to 1\), \(c\to 0\) and hence
\[A=2\left((a+b)^{2}+(b+c)^{2}+(c+a)^{2}\right)\to 0,\]so the left-hand side approaches \(7\).
Thus the sharp constant is
\[C=7.\]Input: 2025.12.08 15:51