Inequality Proof Problems [251-300]
Recommended posting: 【Algebra】 Algebra Index
Restructured the IneqMath training data.
P251. (Turkevici’s inequality) Let $a, b, c, d$ be non-negative real numbers. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c, d \in \mathbb{R}^{+}$:
\[a^{4}+b^{4}+c^{4}+d^{4}+C a b c d \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}+a^{2} c^{2}+b^{2} d^{2}\]S251.1. $C = 2$
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
\[\begin{aligned} &f(a,b,c,d)-f(\sqrt{ac},\,b,\,\sqrt{ac},\,d) \\ =&\ a^{4}+b^{4}+c^{4}+d^{4}+2abcd-a^{2}c^{2}-b^{2}d^{2}-\left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right) \\ &-\Bigl(a^{2}c^{2}+b^{4}+a^{2}c^{2}+d^{4}+2abcd-a^{2}c^{2}-b^{2}d^{2}-2ac\left(b^{2}+d^{2}\right)\Bigr) \\ =&\ a^{4}+c^{4}-2a^{2}c^{2}-\left(b^{2}+d^{2}\right)\left(a^{2}+c^{2}-2ac\right) \\ =&\ (a^{2}-c^{2})^{2}-\left(b^{2}+d^{2}\right)(a-c)^{2} \\ =&\ (a-c)^{2}\Bigl((a+c)^{2}-\left(b^{2}+d^{2}\right)\Bigr)\ge 0. \end{aligned}\]Thus
\[f(a,b,c,d)\ge f(\sqrt{ac},\,b,\,\sqrt{ac},\,d).\]By the SMV theorem we only need to prove that \(f(a,b,c,d)\ge 0\) in the case when \(a=b=c=t\ge d.\)
We have
\[\begin{aligned} f(t,t,t,d)\ge 0 &\Longleftrightarrow 3t^{4}+d^{4}+2t^{3}d\ge 3t^{4}+3t^{2}d^{2} \\ &\Longleftrightarrow d^{4}+2t^{3}d\ge 3t^{2}d^{2}, \end{aligned}\]which immediately follows from \(AM\ge GM.\)
Equality 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.
Therefore, the answer is \(C=2\).
S251.2.
S251.3. \(\frac{3}{2}\,(a^4+b^4+c^4+d^4)\ \ge\ a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2\) can be understood easily (e.g., by the rearrangement inequality). Also, \(6abcd\ \le\ a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2\) can be understood easily (e.g., by the rearrangement inequality). Therefore, the given problem becomes: determine the critical value \(p\) for which \(\frac{6-p}{4}\,(a^4+b^4+c^4+d^4) + p\,abcd \ \ge\ a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2\) is a tight inequality for some \(p\). (This can be seen easily from the fact that when \(p\) is small, the left-hand side is larger, and when \(p\) is large, the right-hand side is larger.) Thus, we can reduce it to the problem of determining that critical value \(p\). If \(a=b=c=d\), we cannot obtain any condition on \(p\). If one of \(a,b,c,d\) is \(0\) and the other three are all equal, we can conclude that \(p\le 2\). Hence, by the existence of a tight inequality, the critical value \(p\) is determined either at a boundary case or at an interior extremal point, so we get \(p=2.\)
S251.3. (Paraphrased)
Let $S_4$ act on $(a,b,c,d)$ by permuting the coordinates. Set
\[s_4:=a^4+b^4+c^4+d^4,\qquad s_2:=\sum_{1\le i<j\le4}x_i^2x_j^2\ ( =a^2b^2+a^2c^2+\cdots+c^2d^2),\qquad t:=abcd,\]where $(x_1,x_2,x_3,x_4)=(a,b,c,d)$. These are $S_4$–invariants (orbit-sums, and $t$ is fixed).
From Cauchy–Schwarz,
\[\Big(\sum_{i=1}^4 x_i^2\Big)^2\le 4\sum_{i=1}^4 x_i^4 \quad\Rightarrow\quad s_4+2s_2\le 4s_4 \quad\Rightarrow\quad s_2\le \frac32,s_4.\]From AM–GM applied to the three perfect matchings of $K_4$ (an $S_4$–orbit),
\[x_1^2x_2^2+x_3^2x_4^2\ge 2t,\quad x_1^2x_3^2+x_2^2x_4^2\ge 2t,\quad x_1^2x_4^2+x_2^2x_3^2\ge 2t,\]and summing gives
\[s_2\ge 6t.\]Consider the $S_4$–invariant one-parameter family
\[\Phi_p:=\frac{6-p}{4}s_4+pt-s_2.\]We want the largest $p$ such that $\Phi_p\ge 0$ for all $a,b,c,d\ge 0$. Since the condition must hold on every $S_4$–orbit, test the boundary orbit representative $(1,1,1,0)$:
\[s_4=3,\quad s_2=3,\quad t=0 \quad\Rightarrow\quad \Phi_p(1,1,1,0)=\frac{6-p}{4}\cdot 3-3\ge 0 \quad\Rightarrow\quad p\le 2.\]If $p=2$ is feasible, then the extremal value is forced:
\[p_{\max}=2.\]Plugging $p=2$ back in yields
\[\frac{6-2}{4}s_4+2t\ge s_2 \quad\Longleftrightarrow\quad s_4+2t\ge s_2,\]i.e. \(a^4+b^4+c^4+d^4+2abcd\ \ge\ a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2,\)
which is Turkevich’s inequality.
P252. Let $P, L, R$ denote the area, perimeter, and circumradius of $\triangle ABC$, respectively. Find the smallest constant $C$ such that the following inequality holds for all triangles $\triangle ABC$:
\[\frac{L P}{R^3} \leq C.\]S252. $C = \frac{27}{4}$
We have
\[\frac{LP}{R^{3}} =\frac{(a+b+c)abc}{R^{3}\cdot 4R} =\frac{2R(\sin\alpha+\sin\beta+\sin\gamma)\cdot 8R^{3}\sin\alpha\sin\beta\sin\gamma}{4R^{4}}.\]i.e.
\[\frac{LP}{R^{3}} =4(\sin\alpha+\sin\beta+\sin\gamma)\sin\alpha\sin\beta\sin\gamma \qquad (1)\]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
\[\frac{\sin\alpha+\sin\beta+\sin\gamma}{3} \le \sin\left(\frac{\alpha+\beta+\gamma}{3}\right) =\sin\left(\frac{\pi}{3}\right) =\frac{\sqrt{3}}{2}.\]Finally from (2) we obtain
\[\frac{LP}{R^{3}} \le \frac{4}{27}\left(\frac{3\sqrt{3}}{2}\right)^{4} =\frac{27}{4}.\]Equality holds when \(a=b=c\), i.e., when the triangle is equilateral. In this case, the maximum value of \(\frac{LP}{R^{3}}\) is achieved, so \(C=\frac{27}{4}\) is the smallest constant for which the inequality always holds.
Therefore, the answer is \(C=\frac{27}{4}\).
P253. Let $a, b, c, d$ be positive real numbers such that $a b c d = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c, d$ satisfying the given constraint:
\[\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+d)}+\frac{1}{d(1+a)} \geq C\]S253. $C = 2$
With the substitutions \(a=\frac{x}{y},\quad b=\frac{t}{x},\quad c=\frac{z}{t},\quad d=\frac{y}{z},\) 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
\[\begin{aligned} \frac{x}{z+t}+\frac{y}{x+t}+\frac{z}{x+y}+\frac{t}{z+y} &=\frac{x^{2}}{xz+xt}+\frac{y^{2}}{yx+yt}+\frac{z^{2}}{zx+zy}+\frac{t^{2}}{tz+ty} \\ &\ge \frac{(x+y+z+t)^{2}}{(xz+xt)+(yx+yt)+(zx+zy)+(tz+ty)} \\ &= \frac{(x+y+z+t)^{2}}{2xz+2yt+xt+xy+yz+tz}. \end{aligned}\]Hence it suffices to prove that
\[\frac{(x+y+z+t)^{2}}{2xz+2yt+xt+xy+yz+tz}\ge 2,\]which is equivalent to
\[(x-z)^{2}+(y-t)^{2}\ge 0.\]Equality occurs when \(x=z\) and \(y=t\), which corresponds to \(a=c=\frac{1}{b}=\frac{1}{d}.\) In this case, the original sum achieves its minimum value. Thus, \(C=2\) is the largest constant for which the inequality always holds.
Therefore, the answer is \(C=2\).
P254. Let $x, y, z$ be real numbers. Find the largest constant $C$ such that the following inequality holds for all $x, y, z \in \mathbb{R}$:
\[x^{4}+y^{4}+z^{4} \geq 4 x y z + C\]S254. $C = -1$
We have
\[\begin{aligned} x^{4}+y^{4}+z^{4}-4xyz+1 &=\left(x^{4}-2x^{2}+1\right)+\left(y^{4}-2y^{2}z^{2}+z^{4}\right)+\left(2y^{2}z^{2}-4xyz+2x^{2}\right) \\ &=\left(x^{2}-1\right)^{2}+\left(y^{2}-z^{2}\right)^{2}+2(yz-x)^{2}\ge 0. \end{aligned}\]so it follows that
\[x^{4}+y^{4}+z^{4}\ge 4xyz-1.\]Equality holds when \(x=1,\ y=z=0\), 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.
Therefore, the answer is \(C=-1\).
P255. Let $a, b, c$ be real numbers different from 1, such that $a+b+c=1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[\frac{1+a^{2}}{1-a^{2}}+\frac{1+b^{2}}{1-b^{2}}+\frac{1+c^{2}}{1-c^{2}} \geq C\]S255. $C = \frac{15}{4}$
Since \(a,b,c>0,\ a\ne 1,\ b\ne 1,\ c\ne 1\) 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
\[\frac{1}{1-a^{2}}+\frac{1}{1-b^{2}}+\frac{1}{1-c^{2}} \ge \frac{9}{(1-a^{2})+(1-b^{2})+(1-c^{2})} =\frac{9}{3-(a^{2}+b^{2}+c^{2})} \ge \frac{9}{3-\frac{1}{3}} =\frac{27}{8} \qquad (3)\]Finally from (2) and (3) we get
\[A\ge \frac{10}{9}\left(\frac{1}{1-a^{2}}+\frac{1}{1-b^{2}}+\frac{1}{1-c^{2}}\right) \ge \frac{10}{9}\cdot \frac{27}{8} =\frac{15}{4},\]with equality iff \(a=b=c=\frac{1}{3}.\)
Equality 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.
Therefore, the answer is \(C=\frac{15}{4}\).
P256. Let $a, b, c \in \mathbb{R}^{+}$. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{abc}{(1+a)(a+b)(b+c)(c+16)} \leq C.\]S256. $C = \frac{1}{81}$
We have
\[\begin{aligned} &(1+a)(a+b)(b+c)(c+16) \\ &=\left(1+\frac{a}{2}+\frac{a}{2}\right) \left(a+\frac{b}{2}+\frac{b}{2}\right) \left(b+\frac{c}{2}+\frac{c}{2}\right) \left(c+8+8\right) \\ &\ge 3\sqrt[3]{\frac{a^{2}}{4}}\cdot 3\sqrt[3]{\frac{ab^{2}}{4}}\cdot 3\sqrt[3]{\frac{bc^{2}}{4}}\cdot 3\sqrt[3]{\frac{64c}{4}} \\ &=81\sqrt[3]{\frac{a^{2}\cdot ab^{2}\cdot bc^{2}\cdot 64c}{4^{4}}} =81\sqrt[3]{\frac{64a^{3}b^{3}c^{3}}{256}} =81\sqrt[3]{\frac{a^{3}b^{3}c^{3}}{4}} =\frac{81}{\sqrt[3]{4}}\,abc \ge 81abc. \end{aligned}\]Thus
\[\frac{abc}{(1+a)(a+b)(b+c)(c+16)}\le \frac{1}{81}.\]Equality holds when, in each AM-GM step, the three summed terms are equal:
\[1=\frac{a}{2}=\frac{a}{2},\quad a=\frac{b}{2}=\frac{b}{2},\quad b=\frac{c}{2}=\frac{c}{2},\quad c=8=8,\]which gives \(a=2,\ b=4,\ c=8.\) In this case, the expression attains its maximum value, so \(C=\frac{1}{81}\) is the minimal constant such that the inequality always holds.
Therefore, the answer is \(C=\frac{1}{81}\).
P257. Let $a, b, c \in (1,2)$ be real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in (1,2)$:
\[\frac{b \sqrt{a}}{4 b \sqrt{c}-c \sqrt{a}}+\frac{c \sqrt{b}}{4 c \sqrt{a}-a \sqrt{b}}+\frac{a \sqrt{c}}{4 a \sqrt{b}-b \sqrt{c}} \geq C\]S257. $C = 1$
Since \(a,b,c\in(1,2)\) we have
\[4b\sqrt{c}-c\sqrt{a}>4\sqrt{c}-2\sqrt{c}=2\sqrt{c}>0.\]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
\[b(a+b+c)\ge a\left(4b\sqrt{c}-c\sqrt{a}\right) \Longleftrightarrow b(a+b+c)\ge 4ab\sqrt{c}-ac\sqrt{a}.\]This is the same as
\[b^{2}+bc+ac\ge 4ab\sqrt{c}-ac\sqrt{a}.\]Using \(a,b,c>0\), we rewrite the left side as
\[b^{2}+bc+ac=(b+c)(a+b)-(ab)\]and applying \(AM\ge GM\) to \(a+b\) and \(b+c\) gives
\[(a+b)(b+c)\ge 4b\sqrt{ac}.\]Hence
\[b(a+b+c)=b^{2}+ab+bc\ge 4b\sqrt{ac}\cdot \frac{a}{\sqrt{a}} = a\left(4b\sqrt{c}-c\sqrt{a}\right),\]so (1) holds.
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 the required result.
Equality holds when \(a=b=c\). In this case each term becomes
\[\frac{b\sqrt{a}}{4b\sqrt{c}-c\sqrt{a}}=\frac{a}{a+a+a}=\frac{1}{3},\]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.
Therefore, the answer is \(C=1\).
P258. Let $a, b, c$ be positive real numbers such that $a b c = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geq C\]S258. $C = \frac{3}{2}$
Without loss of generality we may assume that \(a\ge b\ge c.\) Let \(x=\frac{1}{a},\ y=\frac{1}{b},\ z=\frac{1}{c}.\) 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 \(x\le y\le z.\) Hence
\[x+y\le x+z\le y+z \quad\text{and}\quad \frac{x}{y+z}\le \frac{y}{z+x}\le \frac{z}{x+y}.\]Now by the rearrangement inequality we get
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \ge \frac{xy}{y+z}+\frac{yz}{z+x}+\frac{zx}{x+y},\]and also
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \ge \frac{xz}{y+z}+\frac{yx}{z+x}+\frac{zy}{x+y}.\]So we obtain
\[\begin{aligned} &2\left(\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \\ &\quad=2\left(\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}\right) \\ &\quad\ge \left(\frac{xy}{y+z}+\frac{yz}{z+x}+\frac{zx}{x+y}\right) +\left(\frac{xz}{y+z}+\frac{yx}{z+x}+\frac{zy}{x+y}\right) \\ &\quad= \frac{x(y+z)}{y+z}+\frac{y(z+x)}{z+x}+\frac{z(x+y)}{x+y} \\ &\quad= x+y+z \\ &\quad\ge 3\sqrt[3]{xyz}=3, \end{aligned}\]as required.
Equality holds when \(a=b=c=1\), which implies \(x=y=z=1.\) In this case, the sum becomes
\[\frac{1}{1^{3}(1+1)}+\frac{1}{1^{3}(1+1)}+\frac{1}{1^{3}(1+1)} =3\cdot\frac{1}{2} =\frac{3}{2}.\]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.
Therefore, the answer is \(C=\frac{3}{2}\).
P259. Find the smallest constant $C$ such that for all real numbers $x$, the following inequality holds:
\[2 x^{4} + C \geq 2 x^{3} + x^{2}\]S259. $C = 1$
We have
\[\begin{aligned} 2x^{4}+1-2x^{3}-x^{2} &=1-x^{2}-2x^{3}(1-x) \\ &=(1-x)(1+x)-2x^{3}(1-x) \\ &=(1-x)\left(1+x-2x^{3}\right) \\ &=(1-x)\left(x(1-x^{2})+(1-x^{3})\right) \\ &=(1-x)\left(x(1-x)(1+x)+(1-x)(1+x+x^{2})\right) \\ &=(1-x)^{2}\left(x(1+x)+1+x+x^{2}\right) \\ &=(1-x)^{2}\left((x+1)^{2}+x^{2}\right)\ge 0. \end{aligned}\]Equality occurs if and only if \(x=1\), since then \((1-x)^{2}=0.\) This gives the minimum value of \(C\), so \(C=1\) is the smallest constant for which the inequality always holds.
Therefore, the answer is \(C=1\).
P260. Let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a^{3}+2}{b+2}+\frac{b^{3}+2}{c+2}+\frac{c^{3}+2}{a+2} \geq C.\]S260. $C = 3$
By \(AM\ge GM\) 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
\[\begin{aligned} \frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2} &=\frac{a^{2}}{a(b+2)}+\frac{b^{2}}{b(c+2)}+\frac{c^{2}}{c(a+2)} \\ &\ge \frac{(a+b+c)^{2}}{a(b+2)+b(c+2)+c(a+2)} \\ &=\frac{(a+b+c)^{2}}{ab+bc+ca+2(a+b+c)}. \end{aligned}\]Thus
\[\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2} \ge \frac{(a+b+c)^{2}}{ab+bc+ca+2(a+b+c)} \qquad (2)\]Since
\[(a+b+c)^{2}\ge 3(ab+bc+ca),\]we have
\[ab+bc+ca \le \frac{(a+b+c)^{2}}{3}.\]Hence
\[ab+bc+ca+2(a+b+c)\le \frac{(a+b+c)^{2}}{3}+2(a+b+c).\]So from (2) we get
\[\begin{aligned} \frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2} &\ge \frac{(a+b+c)^{2}}{ab+bc+ca+2(a+b+c)} \\ &\ge \frac{(a+b+c)^{2}}{\frac{(a+b+c)^{2}}{3}+2(a+b+c)} \\ &=\frac{3(a+b+c)^{2}}{(a+b+c)^{2}+6(a+b+c)} =\frac{3(a+b+c)}{(a+b+c)+6}. \end{aligned}\]Thus
\[\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2} \ge \frac{3(a+b+c)}{(a+b+c)+6} \qquad (3)\]Finally by (1), (3) 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 occurs iff \(a=b=c=1.\)
Equality 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 \(3.\)
Therefore, the answer is \(C=3\).
P261. Let $a, b, c$ be the lengths of the sides of a triangle, and let $l_{\alpha}, l_{\beta}, l_{\gamma}$ be the lengths of the bisectors of the respective angles. Let $s$ be the semi-perimeter and $r$ denote the inradius of the triangle. Determine the minimal constant $C$ such that the following inequality holds for all triangles:
\[\frac{l_{\alpha}}{a}+\frac{l_{\beta}}{b}+\frac{l_{\gamma}}{c} \leq C\frac{s}{r}.\]S261. $C = \frac{1}{2}$
The following identities hold:
\[l_{\alpha}=\frac{2\sqrt{bc}}{b+c}\sqrt{s(s-a)},\quad l_{\beta}=\frac{2\sqrt{ca}}{c+a}\sqrt{s(s-b)},\quad l_{\gamma}=\frac{2\sqrt{ab}}{a+b}\sqrt{s(s-c)}.\]From the obvious inequality
\[\frac{2\sqrt{xy}}{x+y}\le 1\]and the previous identities we obtain that
\[l_{\alpha}\le \sqrt{s(s-a)},\quad l_{\beta}\le \sqrt{s(s-b)},\quad l_{\gamma}\le \sqrt{s(s-c)} \qquad (1)\]Also
\[h_{a}\le l_{\alpha},\quad h_{b}\le l_{\beta},\quad h_{c}\le l_{\gamma} \qquad (2)\]So we have
\[\begin{aligned} \frac{l_{\alpha}}{a}+\frac{l_{\beta}}{b}+\frac{l_{\gamma}}{c} &=\frac{l_{\alpha}h_{a}}{2P}+\frac{l_{\beta}h_{b}}{2P}+\frac{l_{\gamma}h_{c}}{2P} \\ &\stackrel{(2)}{\le}\frac{l_{\alpha}^{2}+l_{\beta}^{2}+l_{\gamma}^{2}}{2P} \\ &\stackrel{(1)}{\le}\frac{s(s-a)+s(s-b)+s(s-c)}{2P} \\ &=\frac{3s^{2}-s(a+b+c)}{2P}. \end{aligned}\]Since \(a+b+c=2s\) and \(P=rs\), we get
\[\frac{3s^{2}-s(a+b+c)}{2P} =\frac{3s^{2}-2s^{2}}{2rs} =\frac{s^{2}}{2rs} =\frac{s}{2r}.\]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 all the inequalities above become equalities only in this case. Thus the minimal constant is \(C=\frac{1}{2}.\)
Therefore, the answer is \(C=\frac{1}{2}\).
P262. Let $a, b, c$ be the side lengths of a given triangle. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the triangle inequality:
\[a^{2}+b^{2}+c^{2} < C(a b+b c+c a)\]S262. $C = 2$
Let \(a=x+y,\ b=y+z,\ c=z+x,\quad x,y,z>0.\) Then we have
\[\begin{aligned} &(x+y)^{2}+(y+z)^{2}+(z+x)^{2} \\ &\quad<2\bigl((x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)\bigr), \end{aligned}\]or equivalently
\[2(x^{2}+y^{2}+z^{2})+2(xy+yz+zx)<2(x^{2}+y^{2}+z^{2})+6(xy+yz+zx) \Longleftrightarrow xy+yz+zx>0,\]which is clearly true for \(x,y,z>0.\)
Equality is approached as \(xy+yz+zx\to 0\) (e.g., when one of \(x,y,z\to 0^{+}\)), but it is never achieved for positive \(x,y,z.\) Thus, \(C=2\) is the minimal constant for which the inequality always holds.
Therefore, the answer is \(C=2\).
P263. Let $a, b, c$ be positive real numbers such that $a + b + c = 6$. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[\sqrt[3]{a b + b c} + \sqrt[3]{b c + c a} + \sqrt[3]{c a + a b} \leq C\]S263. $C = 6$
By the power mean inequality we have
\[\frac{\sqrt[3]{ab+bc}+\sqrt[3]{bc+ca}+\sqrt[3]{ca+ab}}{3} \le \sqrt[3]{\frac{(ab+bc)+(bc+ca)+(ca+ab)}{3}},\]i.e.
\[\sqrt[3]{ab+bc}+\sqrt[3]{bc+ca}+\sqrt[3]{ca+ab} \le \sqrt[3]{9(ab+bc+ca)}.\]Since
\[ab+bc+ca\le \frac{(a+b+c)^{2}}{3}=\frac{36}{3}=12,\]we obtain
\[\sqrt[3]{ab+bc}+\sqrt[3]{bc+ca}+\sqrt[3]{ca+ab} \le \sqrt[3]{9\cdot 12} =\sqrt[3]{108} =3\sqrt[3]{4}.\]Equality occurs when \(a=b=c=2\) (so \(a+b+c=6\)), in which case the left side equals
\[3\sqrt[3]{ab+bc}=3\sqrt[3]{8}=6.\]Thus the maximum value of the expression is \(6\), so the minimal constant is \(C=6.\)
Therefore, the answer is \(C=6\).
P264. Let $n \geq 2, n \in \mathbb{N}$ and $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers such that
\[\frac{1}{x_{1}+1998}+\frac{1}{x_{2}+1998}+\cdots+\frac{1}{x_{n}+1998}=\frac{1}{1998}\]Find the largest constant $C$ such that the following inequality holds for all $x_{1}, x_{2}, \ldots, x_{n}$ satisfying the given constraint:
\[\sqrt[n]{x_{1} x_{2} \cdots x_{n}} \geq C\]S264. $C = 1998(n-1)$
After setting
\[\frac{1998}{x_i+1998}=a_i,\qquad i=1,2,\ldots,n,\]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
\[\left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)\cdots\left(\frac{1}{a_n}-1\right)\ge (n-1)^{n}.\]We have
\[\begin{aligned} \frac{1}{a_i}-1 &=\frac{1-a_i}{a_i} =\frac{a_1+\cdots+a_{i-1}+a_{i+1}+\cdots+a_n}{a_i} \\ &\ge \frac{(n-1)\sqrt[n-1]{a_1\cdots a_{i-1}a_{i+1}\cdots a_n}}{a_i}, \end{aligned}\]i.e.
\[\frac{1}{a_i}-1 \ge (n-1)\sqrt[n-1]{\frac{a_1\cdots a_{i-1}a_{i+1}\cdots a_n}{a_i^{\,n-1}}}.\]Multiplying these inequalities for \(i=1,2,\ldots,n\) we obtain
\[\prod_{i=1}^{n}\left(\frac{1}{a_i}-1\right) \ge (n-1)^{n}\sqrt[n-1]{\prod_{i=1}^{n}\frac{a_1\cdots a_{i-1}a_{i+1}\cdots a_n}{a_i^{\,n-1}}} =(n-1)^{n},\]which is the desired inequality.
Equality holds when all \(a_i\) are equal, i.e.
\[a_1=a_2=\cdots=a_n=\frac{1}{n}.\]This corresponds to
\[\frac{1998}{x_i+1998}=\frac{1}{n} \Longleftrightarrow x_i=1998(n-1) \qquad (i=1,2,\ldots,n).\]Thus the minimum value of \(\sqrt[n]{x_1x_2\cdots x_n}\) is achieved at this point, and
\[C=1998(n-1)\]is the largest constant for which the inequality always holds.
Therefore, the answer is \(C=1998(n-1)\).
P265. Let $a, b, c, d$ be positive real numbers such that $abcd = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c, d$ satisfying the given constraint:
\[\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq C\]S265. $C = 1$
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
\[\begin{aligned} \frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}-\frac{1}{1+ab} &=\frac{(1+b)^{2}(1+ab)+(1+a)^{2}(1+ab)-(1+a)^{2}(1+b)^{2}}{(1+a)^{2}(1+b)^{2}(1+ab)} \\ &=\frac{ab(a-b)^{2}+(ab-1)^{2}}{(1+a)^{2}(1+b)^{2}(1+ab)}\ge 0. \end{aligned}\](The second inequality is analogous.)
Equality holds if \(a=b=c=d=1.\)
Equality 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
\[\frac{1}{(1+1)^{2}}=\frac{1}{4},\]so the sum is \(1.\) This gives the minimum value of the sum, so the maximal constant \(C\) is \(1.\)
Therefore, the answer is \(C=1\).
P266. Let $x, y, z$ be real numbers different from 1, such that $x y z = 1$. Find the largest constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint:
\[\left(\frac{3-x}{1-x}\right)^{2}+\left(\frac{3-y}{1-y}\right)^{2}+\left(\frac{3-z}{1-z}\right)^{2} > C\]S266. $C = 7$
Denote
\[A=\left(\frac{3-x}{1-x}\right)^{2}+\left(\frac{3-y}{1-y}\right)^{2}+\left(\frac{3-z}{1-z}\right)^{2}-7.\]We have
\[A=\left(1+\frac{2}{1-x}\right)^{2}+\left(1+\frac{2}{1-y}\right)^{2}+\left(1+\frac{2}{1-z}\right)^{2}-7.\]Let
\[\frac{1}{1-x}=a,\quad \frac{1}{1-y}=b,\quad \frac{1}{1-z}=c.\]Then
\[A=(1+2a)^{2}+(1+2b)^{2}+(1+2c)^{2}-7,\]i.e.
\[A=4a^{2}+4b^{2}+4c^{2}+4a+4b+4c-4.\]Furthermore, the condition \(xyz=1\) is equivalent to
\[(1-x)(1-y)(1-z)=\frac{1}{abc}.\]Since \(1-x=\frac{1}{a}\), \(1-y=\frac{1}{b}\), \(1-z=\frac{1}{c}\) and
\[x=1-\frac{1}{a}=\frac{a-1}{a},\quad y=\frac{b-1}{b},\quad z=\frac{c-1}{c},\]we get
\[xyz=1 \Longleftrightarrow \frac{(a-1)(b-1)(c-1)}{abc}=1 \Longleftrightarrow abc=(a-1)(b-1)(c-1),\]which expands to
\[ab+bc+ca=a+b+c-1.\]Using this, we obtain
\[\begin{aligned} A &=4(a^{2}+b^{2}+c^{2})+4(a+b+c)-4 \\ &=4(a^{2}+b^{2}+c^{2})+4(ab+bc+ca) \\ &=2\left((a+b)^{2}+(b+c)^{2}+(c+a)^{2}\right)\ge 0. \end{aligned}\]Hence
\[\left(\frac{3-x}{1-x}\right)^{2}+\left(\frac{3-y}{1-y}\right)^{2}+\left(\frac{3-z}{1-z}\right)^{2}\ge 7.\]Equality would require \(a+b=0,\ b+c=0,\ c+a=0,\) hence \(a=b=c=0,\) which is impossible. Therefore, the inequality is strict, but the largest constant that always works is still \(C=7\).
Therefore, the answer is \(C=7\).
P267. Let $a, b \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all $a, b \in \mathbb{R}^{+}$:
\[C(a^3 + b^3) \geq (a + b)^3\]S267. $C = 4$
The function \(f(x)=x^{3}\) is convex on \((0,+\infty),\) thus from Jensen’s inequality it follows that
\[\left(\frac{a+b}{2}\right)^{3}\le \frac{a^{3}+b^{3}}{2} \Longleftrightarrow 4(a^{3}+b^{3})\ge (a+b)^{3}.\]Equality holds when \(a=b,\) since then \(a^{3}+b^{3}=2a^{3}\) and \((a+b)^{3}=(2a)^{3}=8a^{3},\) so \(4(a^{3}+b^{3})=8a^{3}=(a+b)^{3}.\) This gives the minimum value of \(C,\) so \(C=4\) is the smallest constant for which the inequality always holds.
Therefore, the answer is \(C=4\).
P268. Let $x, y, z > 0$ be real numbers. Determine the maximal constant $C$ such that the following inequality holds for all $x, y, z$:
\[(x y + y z + z x) \left( \frac{1}{(x + y)^2} + \frac{1}{(y + z)^2} + \frac{1}{(z + x)^2} \right) \geq C.\]S268. $C = \frac{9}{4}$
The given inequality is equivalent to
\[4(xy+yz+zx)\Bigl((z+x)^{2}(y+z)^{2}+(x+y)^{2}(z+x)^{2}+(x+y)^{2}(y+z)^{2}\Bigr) \ge 9(x+y)^{2}(y+z)^{2}(z+x)^{2}.\]Let us denote
\[p=x+y+z,\quad q=xy+yz+zx,\quad r=xyz.\]By \(I_{5}\) and \(I_{7}\) we have
\[(x+y)^{2}(y+z)^{2}(z+x)^{2}=(pq-r)^{2},\]and
\[(x+y)^{2}(y+z)^{2}+(y+z)^{2}(z+x)^{2}+(z+x)^{2}(x+y)^{2} =(p^{2}+q)^{2}-4p(pq-r).\]So we can rewrite the inequality as follows:
\[\begin{aligned} &4q\Bigl((p^{2}+q)^{2}-4p(pq-r)\Bigr)\ge 9(pq-r)^{2} \\ &\Longleftrightarrow\ 4p^{4}q-17p^{2}q^{2}+4q^{3}+34pqr-9r^{2}\ge 0 \\ &\Longleftrightarrow\ 3pq\left(p^{3}-4pq+9r\right) +q\left(p^{4}-5p^{2}q+4q^{2}+6pr\right) +r(pq-9r)\ge 0. \end{aligned}\]The last inequality follows from \(N_{1},N_{2},N_{3}\) and the fact that \(p,q,r>0.\) Equality occurs if and only if \(x=y=z.\)
Equality holds when \(x=y=z,\) in which case
\[xy+yz+zx=3x^{2} \quad\text{and}\quad \frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}} =\frac{3}{(2x)^{2}}=\frac{3}{4x^{2}},\]so the left side equals
\[3x^{2}\cdot \frac{3}{4x^{2}}=\frac{9}{4}.\]Thus, the minimum value of the expression is \(C=\frac{9}{4}.\)
Therefore, the answer is \(C=\frac{9}{4}\).
P269. Let $a, b, c$ be the lengths of the sides of a triangle. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$:
\[(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c} \leq C (a^{a} b^{b} c^{c}).\]S269. $C = 1$
By the weighted power mean inequality we have
\[\begin{aligned} \sqrt[a+b+c]{\left(\frac{a+b-c}{a}\right)^{a}\left(\frac{b+c-a}{b}\right)^{b}\left(\frac{c+a-b}{c}\right)^{c}} &\le \frac{1}{a+b+c}\left(a\cdot\frac{a+b-c}{a}+b\cdot\frac{b+c-a}{b}+c\cdot\frac{c+a-b}{c}\right) \\ &=\frac{(a+b-c)+(b+c-a)+(c+a-b)}{a+b+c} =1. \end{aligned}\]i.e.
\[(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c}\le a^{a}b^{b}c^{c}.\]Equality occurs when
\[\frac{a+b-c}{a}=\frac{b+c-a}{b}=\frac{c+a-b}{c},\]which is equivalent to \(a=b=c\) (the equilateral triangle case). Hence the minimal constant is \(C=1.\)
Therefore, the answer is \(C=1\).
P270. Let $x, y, z \in \mathbb{R}^{+}$ such that $x y z = 1$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq x + y + z$. Determine the largest constant $C$ such that for any natural number $n$, the following inequality holds for all $x, y, z$:
\[\frac{1}{x^{n}} + \frac{1}{y^{n}} + \frac{1}{z^{n}} \geq C (x^{n} + y^{n} + z^{n}).\]S270. $C = 1$
After setting
\[x=\frac{a}{b},\quad y=\frac{b}{c},\quad z=\frac{c}{a},\]the initial condition
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge x+y+z\]becomes
\[\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\ge \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \Longleftrightarrow a^{2}b+b^{2}c+c^{2}a\ge ab^{2}+bc^{2}+ca^{2} \Longleftrightarrow (a-b)(b-c)(c-a)\le 0.\]Let \(n\in\mathbb{N}\), and take
\[A=a^{n},\quad B=b^{n},\quad C=c^{n}.\]Then \(a\ge b\Longleftrightarrow A\ge B\) and \(a\le b\Longleftrightarrow A\le B\), etc. Hence
\[(a-b)(b-c)(c-a)\le 0 \Longleftrightarrow (A-B)(B-C)(C-A)\le 0.\]But
\[(A-B)(B-C)(C-A)\le 0 \Longleftrightarrow A^{2}B+B^{2}C+C^{2}A\ge AB^{2}+BC^{2}+CA^{2} \Longleftrightarrow \frac{B}{A}+\frac{C}{B}+\frac{A}{C}\ge \frac{A}{B}+\frac{B}{C}+\frac{C}{A}.\]Since
\[\frac{B}{A}=\left(\frac{b}{a}\right)^{n}=\frac{1}{x^{n}},\quad \frac{C}{B}=\left(\frac{c}{b}\right)^{n}=\frac{1}{y^{n}},\quad \frac{A}{C}=\left(\frac{a}{c}\right)^{n}=\frac{1}{z^{n}},\]and
\[\frac{A}{B}=x^{n},\quad \frac{B}{C}=y^{n},\quad \frac{C}{A}=z^{n},\]we obtain
\[\frac{1}{x^{n}}+\frac{1}{y^{n}}+\frac{1}{z^{n}}\ge x^{n}+y^{n}+z^{n}.\]Equality holds when \(a=b=c,\) which corresponds to \(x=y=z=1.\) In this case,
\[\frac{1}{x^{n}}+\frac{1}{y^{n}}+\frac{1}{z^{n}}=3 \quad\text{and}\quad x^{n}+y^{n}+z^{n}=3,\]so the inequality becomes \(3\ge C\cdot 3,\) i.e. \(C=1.\) Thus the maximal constant is \(C=1.\)
Therefore, the answer is \(C=1\).
P271. Let $x, y, z \in \mathbb{R}^{+}$ such that $x + y + z = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $x, y, z$:
\[\frac{1}{1-x y}+\frac{1}{1-y z}+\frac{1}{1-z x} \leq C.\]S271. $C = \frac{27}{8}$
Let \(p=x+y+z=1,\quad q=xy+yz+zx,\quad r=xyz.\) It can easily be shown that
\[(1-xy)(1-yz)(1-zx)=1-q+pr-r^{2},\]and
\[(1-xy)(1-yz)+(1-yz)(1-zx)+(1-zx)(1-xy)=3-2q+pr.\]So the given inequality becomes
\[\begin{aligned} 8(3-2q+pr)&\le 27(1-q+pr-r^{2}) \\ \Longleftrightarrow\quad 3-11q+19pr-27r^{2}&\ge 0. \end{aligned}\]Since \(p=1,\) we need to show that
\[3-11q+19r-27r^{2}\ge 0.\]By \(N_{5}: p^{3}\ge 27r\) we have \(1\ge 27r,\) i.e. \(r\ge 27r^{2}.\) Therefore
\[3-11q+19r-27r^{2}\ge 3-11q+19r-r=3-11q+18r.\]So it suffices to prove that
\[3-11q+18r\ge 0.\]We have
\[\begin{aligned} 3-11q+18r\ge 0 &\Longleftrightarrow 3-11(xy+yz+zx)+18xyz\ge 0 \\ &\Longleftrightarrow 11(xy+yz+zx)-18xyz\le 3. \end{aligned}\]Applying \(AM\ge GM\) we deduce
\[\begin{aligned} 11(xy+yz+zx)-18xyz &=xy(11-18z)+11z(x+y) \\ &\le \frac{(x+y)^{2}}{4}(11-18z)+11z(x+y) \\ &=\frac{(1-z)^{2}}{4}(11-18z)+11z(1-z) \\ &=\frac{(1-z)\left((1-z)(11-18z)+44z\right)}{4} \\ &=\frac{11+4z+3z^{2}-18z^{3}}{4}. \end{aligned}\]So it remains to show that
\[\frac{11+4z+3z^{2}-18z^{3}}{4}\le 3 \Longleftrightarrow 4z+3z^{2}-18z^{3}\le 1 \Longleftrightarrow 18z^{3}-3z^{2}-4z+1\ge 0 \Longleftrightarrow (3z-1)^{2}(2z+1)\ge 0,\]which is obvious.
Equality holds when \(x=y=z=\frac{1}{3},\) since then \(x+y+z=1\) and \(xy=yz=zx=\frac{1}{9},\) so
\[\frac{1}{1-xy}=\frac{1}{1-\frac{1}{9}}=\frac{9}{8},\]and the sum is \(3\cdot \frac{9}{8}=\frac{27}{8}.\) This gives the maximum value of the sum, so \(C=\frac{27}{8}\) is the minimal constant such that the inequality always holds.
Therefore, the answer is \(C=\frac{27}{8}\).
P272. Let $\alpha_{i}>0, i=1,2, \ldots, n$, be real numbers such that $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=1$. Determine the maximal constant $C$ such that the following inequality holds for all $\alpha_{i}$:
\[\alpha_{1}^{\alpha_{1}} \alpha_{2}^{\alpha_{2}} \cdots \alpha_{n}^{\alpha_{n}} \geq C.\]S272. $C = \frac{1}{n}$
If we take
\[a_i=\frac{1}{\alpha_i},\qquad i=1,2,\ldots,n,\]then by the weighted \(AM\text{-}GM\) inequality we get
\[\left(\frac{1}{\alpha_1}\right)^{\alpha_1} \left(\frac{1}{\alpha_2}\right)^{\alpha_2} \cdots \left(\frac{1}{\alpha_n}\right)^{\alpha_n} \le \alpha_1\cdot\frac{1}{\alpha_1} +\alpha_2\cdot\frac{1}{\alpha_2} +\cdots +\alpha_n\cdot\frac{1}{\alpha_n} =n.\]i.e.
\[\frac{1}{\alpha_1^{\alpha_1}\alpha_2^{\alpha_2}\cdots\alpha_n^{\alpha_n}}\le n \Longleftrightarrow \alpha_1^{\alpha_1}\alpha_2^{\alpha_2}\cdots\alpha_n^{\alpha_n}\ge \frac{1}{n}.\]Equality holds when
\[\alpha_1=\alpha_2=\cdots=\alpha_n=\frac{1}{n},\]in which case
\[\alpha_1^{\alpha_1}\cdots\alpha_n^{\alpha_n} =\left(\frac{1}{n}\right)^{\frac{1}{n}+\cdots+\frac{1}{n}} =\left(\frac{1}{n}\right)^{1} =\frac{1}{n}.\]Thus the minimum value of the product is \(\frac{1}{n},\) so the maximal constant is \(C=\frac{1}{n}.\)
Therefore, the answer is \(C=\frac{1}{n}\).
P273. Let $a, b \in \mathbb{R}^{+}$ and $n \in \mathbb{N}$. Find the smallest constant $C$ such that the following inequality holds for all $a, b \in \mathbb{R}^{+}$ and $n \in \mathbb{N}$:
\[(a+b)^{n}\left(a^{n}+b^{n}\right) \leq C\left(a^{2 n}+b^{2 n}\right)\]S273. $C = 2^n$
By the power mean inequality, for any \(x,y\in\mathbb{R}^{+}\) and \(n\in\mathbb{N}\), we have
\[\left(\frac{x+y}{2}\right)^{n}\le \frac{x^{n}+y^{n}}{2}.\]Therefore
\[\begin{aligned} (a+b)^{n}\left(a^{n}+b^{n}\right) &=2^{n}\left(\frac{a+b}{2}\right)^{n}\left(a^{n}+b^{n}\right) \\ &\le 2^{n}\left(\frac{a^{n}+b^{n}}{2}\right)\left(a^{n}+b^{n}\right) \\ &=2^{n}\cdot \frac{\left(a^{n}+b^{n}\right)^{2}}{2} \\ &\le 2^{n}\cdot \frac{2\left(a^{2n}+b^{2n}\right)}{2} \\ &=2^{n}\left(a^{2n}+b^{2n}\right). \end{aligned}\]Equality holds when \(a=b\), since all the inequalities above become equalities in this case. Thus the smallest constant is \(C=2^{n}.\)
Therefore, the answer is \(C=2^{n}\).
P274. Let $a_{1}, a_{2}, \ldots, a_{n} \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$:
\[\sum_{k=1}^{n} k a_{k} \leq C + \sum_{k=1}^{n} a_{k}^{k}\]S274. $C = \binom{n}{2}$
For \(1\le k\le n\) we have
\[a_{k}^{k}+(k-1)=a_{k}^{k}+\underbrace{1+1+\cdots+1}_{k-1} \ge k\sqrt[k]{a_{k}^{k}\cdot \underbrace{1\cdot 1\cdots 1}_{k-1}} =ka_{k}.\]After adding these inequalities for \(1\le k\le n\) we get
\[\sum_{k=1}^{n}ka_{k} \le \sum_{k=1}^{n}a_{k}^{k}+\sum_{k=1}^{n}(k-1) =\sum_{k=1}^{n}a_{k}^{k}+\frac{n(n-1)}{2} =\sum_{k=1}^{n}a_{k}^{k}+\binom{n}{2}.\]Equality holds when \(a_{k}=1\) for all \(k\), since then \(a_{k}^{k}=1\) and \(ka_{k}=k\), so both sides are equal. Thus the smallest constant is
\[C=\binom{n}{2}.\]Therefore, the answer is \(C=\binom{n}{2}\).
P275. Let $a, b, c$ be positive real numbers such that $a + b + c \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[a + b + c \geq \frac{C}{a+b+c} + \frac{2}{abc}.\]S275. $C = 3$
By \(AM\ge HM\) we get
\[a+b+c \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9}{a+b+c},\]i.e.
\[\frac{a+b+c}{3}\ge \frac{3}{a+b+c}\qquad (1)\]We will prove that
\[\frac{2(a+b+c)}{3}\ge \frac{2}{abc}\qquad (2)\]i.e.
\[a+b+c\ge \frac{3}{abc}.\]Using the well-known inequality
\[(xy+yz+zx)^{2}\ge 3(xy+yz+zx),\]we obtain
\[(a+b+c)^{2}\ge \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2} \ge 3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right) =3\frac{a+b+c}{abc},\]i.e.
\[a+b+c\ge \frac{3}{abc}.\]After adding (1) and (2) we get the required inequality.
Equality holds when \(a=b=c=1,\) since then \(a+b+c=3,\) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3,\) and \(abc=1,\) so the original constraint is satisfied with equality, and the inequality becomes
\[3\ge \frac{3}{3}+2=3.\]Thus the maximum value of \(C\) for which the inequality always holds is \(C=3.\)
Therefore, the answer is \(C=3\).
P276. Let $a, b, c, x, y, z$ be positive real numbers such that $x + y + z = 1$. Determine the maximal constant $C$ such that the following inequality holds for all $a, b, c, x, y, z$:
\[a x + b y + c z + C \sqrt{(x y + y z + z x)(a b + b c + c a)} \leq a+b+c.\]S276.1. $C = 2$
The inequality being homogeneous in \(a,b,c\), we can assume that \(a+b+c=1.\) We apply the AM-GM inequality and we find that
\[2\sqrt{(xy+yz+zx)(ab+bc+ca)}\le (xy+yz+zx)+(ab+bc+ca).\]Hence
\[ax+by+cz+2\sqrt{(xy+yz+zx)(ab+bc+ca)} \le ax+by+cz+xy+yz+zx+ab+bc+ca.\]Consequently, it is enough to show that
\[xy+yz+zx+ab+bc+ca\le 1-ax-by-cz.\]But since
\[xy+yz+zx=\frac{(x+y+z)^{2}-(x^{2}+y^{2}+z^{2})}{2} =\frac{1-x^{2}-y^{2}-z^{2}}{2},\]and similarly
\[ab+bc+ca=\frac{1-a^{2}-b^{2}-c^{2}}{2},\]we have
\[xy+yz+zx+ab+bc+ca =\frac{1-x^{2}-y^{2}-z^{2}}{2}+\frac{1-a^{2}-b^{2}-c^{2}}{2}.\]Thus the desired inequality is equivalent to
\[\frac{1-x^{2}-y^{2}-z^{2}}{2}+\frac{1-a^{2}-b^{2}-c^{2}}{2}\le 1-ax-by-cz,\]which is the same as
\[(x-a)^{2}+(y-b)^{2}+(z-c)^{2}\ge 0.\]Equality holds when \(x=a,\ y=b,\ z=c.\) In this case, the inequality becomes an equality, and \(C=2\) is the maximal value for which the inequality always holds.
Therefore, the answer is \(C=2\).
S276.2. $C = 2$
We will use the Cauchy-Schwarz inequality twice. First, we can write
\[ax+by+cz\le \sqrt{a^{2}+b^{2}+c^{2}}\cdot \sqrt{x^{2}+y^{2}+z^{2}}.\]Also,
\[2\sqrt{(xy+yz+zx)(ab+bc+ca)} =\sqrt{2(ab+bc+ca)}\cdot \sqrt{2(xy+yz+zx)}.\]Therefore,
\[\begin{aligned} &ax+by+cz+2\sqrt{(xy+yz+zx)(ab+bc+ca)} \\ &\le \sqrt{a^{2}+b^{2}+c^{2}}\cdot \sqrt{x^{2}+y^{2}+z^{2}} +\sqrt{2(ab+bc+ca)}\cdot \sqrt{2(xy+yz+zx)} \\ &\le \sqrt{a^{2}+b^{2}+c^{2}+2(ab+bc+ca)}\cdot \sqrt{x^{2}+y^{2}+z^{2}+2(xy+yz+zx)} \\ &=\sqrt{(a+b+c)^{2}}\cdot \sqrt{(x+y+z)^{2}} =(a+b+c)(x+y+z). \end{aligned}\]If \(a+b+c=1\) and \(x+y+z=1\), then the right-hand side equals \(1.\)
Equality holds when \(a=b=c\) and \(x=y=z.\) In this case, the inequality becomes an equality, and \(C=2\) is the maximal value for which the inequality always holds.
Therefore, the answer is \(C=2\).
P277. Let $a, b, c$ be positive real numbers such that $ab + bc + ca = 3$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[\left(a^{3}-a+5\right)\left(b^{5}-b^{3}+5\right)\left(c^{7}-c^{5}+5\right) \geq C\]S277. $C = 125$
For any real number \(x,\) the numbers \(x-1,\ x^{2}-1,\ x^{3}-1\) and \(x^{5}-1\) are of the same sign.
Therefore
\[(x-1)(x^{2}-1)\ge 0,\quad (x^{2}-1)(x^{3}-1)\ge 0,\quad (x^{2}-1)(x^{5}-1)\ge 0,\]i.e.
\[\begin{aligned} a^{3}-a^{2}-a+1&\ge 0, \\ b^{5}-b^{3}-b^{2}+1&\ge 0, \\ c^{7}-c^{5}-c^{2}+1&\ge 0. \end{aligned}\]So it follows that
\[a^{3}-a+5\ge a^{2}+4,\quad b^{5}-b^{3}+5\ge b^{2}+4,\quad c^{7}-c^{5}+5\ge c^{2}+4.\]Multiplying these inequalities gives
\[(a^{3}-a+5)(b^{5}-b^{3}+5)(c^{7}-c^{5}+5) \ge (a^{2}+4)(b^{2}+4)(c^{2}+4) \qquad (1)\]We’ll prove that
\[(a^{2}+4)(b^{2}+4)(c^{2}+4)\ge 25(ab+bc+ca+2) \qquad (2)\]We have
\[\begin{aligned} (a^{2}+4)(b^{2}+4)(c^{2}+4) &=a^{2}b^{2}c^{2}+4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})+16(a^{2}+b^{2}+c^{2})+64 \\ &=\Bigl(a^{2}b^{2}c^{2}+(a^{2}+b^{2}+c^{2})+2\Bigr) +4\Bigl(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+3\Bigr) \\ &\quad +15(a^{2}+b^{2}+c^{2})+50. \end{aligned}\]Thus
\[(a^{2}+4)(b^{2}+4)(c^{2}+4) =\Bigl(a^{2}b^{2}c^{2}+(a^{2}+b^{2}+c^{2})+2\Bigr) +4\Bigl(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+3\Bigr) +15(a^{2}+b^{2}+c^{2})+50 \qquad (3)\]By the obvious inequalities
\[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\ge 0 \quad\text{and}\quad (ab-1)^{2}+(bc-1)^{2}+(ca-1)^{2}\ge 0,\]we obtain
\[a^{2}+b^{2}+c^{2}\ge ab+bc+ca \qquad (4)\]and
\[a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+3\ge 2(ab+bc+ca). \qquad (5)\]We’ll prove that
\[a^{2}b^{2}c^{2}+(a^{2}+b^{2}+c^{2})+2\ge 2(ab+bc+ca). \qquad (6)\]Lemma. Let \(x,y,z>0.\) Then
\[3xyz+x^{3}+y^{3}+z^{3}\ge 2\left((xy)^{3/2}+(yz)^{3/2}+(zx)^{3/2}\right).\]Proof. By Schur’s inequality and \(AM\ge GM\) we have
\[\begin{aligned} x^{3}+y^{3}+z^{3}+3xyz &\ge (x^{2}y+y^{2}x)+(y^{2}z+z^{2}y)+(z^{2}x+x^{2}z) \\ &\ge 2\left((xy)^{3/2}+(yz)^{3/2}+(zx)^{3/2}\right). \end{aligned}\]By the lemma for \(x=a^{2/3},\ y=b^{2/3},\ z=c^{2/3}\) we deduce
\[3(abc)^{2/3}+a^{2}+b^{2}+c^{2}\ge 2(ab+bc+ca).\]Therefore it suffices to prove that
\[a^{2}b^{2}c^{2}+2\ge 3(abc)^{2/3},\]which follows immediately by \(AM\ge GM\).
Thus we have proved inequality (6). Now by (3), (4), (5) and (6) we obtain inequality (2).
Finally by (1), (2) and since \(ab+bc+ca=3\) we obtain the required inequality. Equality occurs if and only if \(a=b=c=1.\)
Equality holds when \(a=b=c=1,\) which satisfies \(ab+bc+ca=3.\) In this case, the expression evaluates to
\[(1^{3}-1+5)(1^{5}-1^{3}+5)(1^{7}-1^{5}+5)=5\cdot 5\cdot 5=125.\]This gives the minimum value of the expression, so \(C=125\) is the largest constant for which the inequality always holds.
Therefore, the answer is \(C=125\).
P278. Let $n$ be an integer greater than 2. Find the greatest real number $C_{\text{min}}$ and the least real number $C_{\text{max}}$ such that for any positive real numbers $x_1, x_2, \ldots, x_n$ (with $x_n = x_0$ and $x_{n+1} = x_1$), the following inequality holds:
\[C_{\text{min}} \leq \sum_{i=1}^{n} \frac{x_i}{x_{i-1} + 2(n-1)x_i + x_{i+1}} \leq C_{\text{max}}\]S278. \(C_{\text{min}}=\frac{1}{2(n-1)},\quad C_{\text{max}}=\frac{1}{2}.\)
We will prove that \(m_{n}=\frac{1}{2(n-1)},\quad M_{n}=\frac{1}{2}.\)
First, let us see that the inequality
\[\sum_{i=1}^{n}\frac{x_{i}}{x_{i-1}+2(n-1)x_{i}+x_{i+1}}\ge \frac{1}{2(n-1)}\]is trivial, because
\[x_{i-1}+2(n-1)x_{i}+x_{i+1}\le 2(n-1)\sum_{k=1}^{n}x_{k} \quad\text{for all}\quad i.\]This shows that
\[m_{n}\ge \frac{1}{2(n-1)}.\]Taking \(x_{i}=x^{i}\), the expression becomes
\[\frac{1}{x+x^{n-1}+2(n-1)} +\frac{(n-2)x}{1+2(n-1)x+x^{2}} +\frac{x^{n-1}}{1+2(n-1)x^{n-1}+x^{n-2}},\]and taking the limit when \(x\to 0\), we find that
\[m_{n}\le \frac{1}{2(n-1)}.\]Thus
\[m_{n}=\frac{1}{2(n-1)}.\]Now, we will prove that \(M_{n}\le \frac{1}{2}.\) Of course, it suffices to prove that for any \(x_{1},x_{2},\ldots,x_{n}>0\) we have
\[\sum_{i=1}^{n}\frac{x_{i}}{x_{i-1}+2(n-1)x_{i}+x_{i+1}}\le \frac{1}{2}.\]But it is clear that
\[\begin{aligned} \sum_{i=1}^{n}\frac{2x_{i}}{x_{i-1}+2(n-1)x_{i}+x_{i+1}} &\le \sum_{i=1}^{n}\frac{2x_{i}}{2\sqrt{x_{i-1}x_{i+1}}+2(n-1)x_{i}} \\ &=\sum_{i=1}^{n}\frac{1}{(n-1)+\frac{\sqrt{x_{i-1}x_{i+1}}}{x_{i}}}. \end{aligned}\]Taking \(\frac{\sqrt{x_{i-1}x_{i+1}}}{x_{i}}=a_{i},\) we have to prove that if
\[\prod_{i=1}^{n}a_{i}=1,\]then
\[\sum_{i=1}^{n}\frac{1}{n-1+a_{i}}\le 1.\]But this has already been proved in Problem 84. Hence
\[\sum_{i=1}^{n}\frac{2x_{i}}{x_{i-1}+2(n-1)x_{i}+x_{i+1}}\le 1,\]which yields
\[\sum_{i=1}^{n}\frac{x_{i}}{x_{i-1}+2(n-1)x_{i}+x_{i+1}}\le \frac{1}{2}.\]Thus
\[M_{n}\le \frac{1}{2}.\]Because for \(x_{1}=x_{2}=\cdots=x_{n}\) we have equality, we deduce that
\[M_{n}=\frac{1}{2},\]which solves the problem.
Equality in the lower bound \(C_{\text{min}}=\frac{1}{2(n-1)}\) is achieved in the limit as one variable tends to zero and the others are fixed, for example by taking
\[x_{1}=x_{2}=\cdots=x_{n-1}=1,\quad x_{n}\to 0.\]Equality in the upper bound \(C_{\text{max}}=\frac{1}{2}\) is achieved when all variables are equal, i.e.
\[x_{1}=x_{2}=\cdots=x_{n}.\]Therefore, the answer is
\[C_{\text{min}}=\frac{1}{2(n-1)},\quad C_{\text{max}}=\frac{1}{2}.\]P279. Let $x_{1}, x_{2}, \ldots, x_{n}$ be non-negative real numbers such that $x_{1} + x_{2} + \cdots + x_{n} \leq \frac{1}{2}$. Find the largest constant $C$ such that the following inequality holds for all $x_{1}, x_{2}, \ldots, x_{n}$ satisfying the given constraint:
\[\left(1 - x_{1}\right)\left(1 - x_{2}\right) \cdots \left(1 - x_{n}\right) \geq C\]S279. $C = \frac{1}{2}$
From \(x_{1}+x_{2}+\cdots+x_{n}\le \frac{1}{2}\) and the fact that \(x_{1},x_{2},\ldots,x_{n}\) are nonnegative we deduce that
\[0\le x_{i}\le \frac{1}{2}<1, \quad\text{i.e.}\quad -x_{i}>-1, \quad\text{for all}\quad i=1,2,\ldots,n.\]Hence all \(-x_{i}\) are of the same sign.
Applying Bernoulli’s inequality we obtain
\[\begin{aligned} (1-x_{1})(1-x_{2})\cdots(1-x_{n}) &=\left(1+(-x_{1})\right)\left(1+(-x_{2})\right)\cdots\left(1+(-x_{n})\right) \\ &\ge 1+\left((-x_{1})+(-x_{2})+\cdots+(-x_{n})\right) \\ &=1-(x_{1}+x_{2}+\cdots+x_{n}) \\ &\ge 1-\frac{1}{2}=\frac{1}{2}. \end{aligned}\]Equality holds when \(x_{1}+x_{2}+\cdots+x_{n}=\frac{1}{2}\) and all but one of the variables are zero, for example
\[x_{1}=\frac{1}{2},\quad x_{2}=x_{3}=\cdots=x_{n}=0.\]Thus the minimum value of the product is \(\frac{1}{2}\), so the largest constant \(C\) for which the inequality always holds is
\[C=\frac{1}{2}.\]Therefore, the answer is \(C=\frac{1}{2}\).
P280. Let $a, b, c \in (-3, 3)$ such that $\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c}=\frac{1}{3-a}+\frac{1}{3-b}+\frac{1}{3-c}$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given condition:
\[\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c} \geq C\]S280. $C = 1$
By the inequality \(AM\ge HM\) we have
\[\bigl((3+a)+(3+b)+(3+c)\bigr)\left(\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c}\right)\ge 9 \qquad (1)\]and
\[\bigl((3-a)+(3-b)+(3-c)\bigr)\left(\frac{1}{3-a}+\frac{1}{3-b}+\frac{1}{3-c}\right)\ge 9.\]If we use the condition \(a+b+c=0\), then
\[(3-a)+(3-b)+(3-c)=9-(a+b+c)=9,\]so the previous inequality becomes
\[9\left(\frac{1}{3-a}+\frac{1}{3-b}+\frac{1}{3-c}\right)\ge 9, \qquad\text{i.e.}\qquad \frac{1}{3-a}+\frac{1}{3-b}+\frac{1}{3-c}\ge 1 \qquad (2)\]After adding (1) and (2) we obtain
\[9\left(\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c}\right) +9\left(\frac{1}{3-a}+\frac{1}{3-b}+\frac{1}{3-c}\right) \ge 18.\]In particular,
\[\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c}\ge 1.\]Equality holds when \(a=b=c=0,\) since then
\[\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c} =3\cdot \frac{1}{3}=1.\]Thus the minimum value of the sum is \(1,\) so the largest constant \(C\) for which the inequality always holds is
\[C=1.\]Therefore, the answer is \(C=1\).
P281. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[\frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq C\]S281. $C = 1$
Applying \(AM\ge GM\) gives us
\[\frac{a^{2}}{a+2b^{3}} =a-\frac{2ab^{3}}{a+2b^{3}} \ge a-\frac{2ab^{3}}{3\sqrt[3]{ab^{4}}} =a-\frac{2ba^{2/3}}{3}.\]Analogously,
\[\frac{b^{2}}{b+2c^{3}}\ge b-\frac{2cb^{2/3}}{3}, \quad\text{and}\quad \frac{c^{2}}{c+2a^{3}}\ge c-\frac{2ac^{2/3}}{3}.\]Adding these three inequalities implies
\[\frac{a^{2}}{a+2b^{3}}+\frac{b^{2}}{b+2c^{3}}+\frac{c^{2}}{c+2a^{3}} \ge (a+b+c)-\frac{2}{3}\left(ba^{2/3}+cb^{2/3}+ac^{2/3}\right).\]So it is enough to prove that
\[(a+b+c)-\frac{2}{3}\left(ba^{2/3}+cb^{2/3}+ac^{2/3}\right)\ge 1,\]i.e.
\[ba^{2/3}+cb^{2/3}+ac^{2/3}\le 3 \qquad (1)\]After another application of \(AM\ge GM\) we get
\[a^{2/3}\le \frac{2a+1}{3},\quad b^{2/3}\le \frac{2b+1}{3},\quad c^{2/3}\le \frac{2c+1}{3}.\]Multiplying by \(b,c,a\) respectively and summing yields
\[\begin{aligned} ba^{2/3}+cb^{2/3}+ac^{2/3} &\le \frac{b(2a+1)+c(2b+1)+a(2c+1)}{3} \\ &=\frac{a+b+c+2(ab+bc+ca)}{3}. \end{aligned}\]Using
\[ab+bc+ca\le \frac{(a+b+c)^{2}}{3},\]we obtain
\[ba^{2/3}+cb^{2/3}+ac^{2/3} \le \frac{(a+b+c)+\frac{2}{3}(a+b+c)^{2}}{3}.\]For \(a+b+c=3\) this gives
\[ba^{2/3}+cb^{2/3}+ac^{2/3}\le \frac{3+\frac{2}{3}\cdot 9}{3}=3,\]i.e. we have proved (1), and we are done.
Equality holds when \(a=b=c=1,\) since then \(a+b+c=3\) and all terms are equal, so the minimum value of the sum is achieved and \(C=1\) is the largest constant for which the inequality always holds.
Therefore, the answer is \(C=1\).
P282. Let $t_{a}, t_{b}, t_{c}$ be the lengths of the medians, and $a, b, c$ be the lengths of the sides of a given triangle. Find the smallest constant $C$ such that the following inequality holds for all triangles:
\[t_{a} t_{b}+t_{b} t_{c}+t_{c} t_{a} < C(a b+b c+c a)\]S282. $C = \frac{5}{4}$
We can easily show the inequalities
\[t_{a}<\frac{b+c}{2},\quad t_{b}<\frac{a+c}{2},\quad t_{c}<\frac{a+b}{2}.\]After adding these we get
\[t_{a}+t_{b}+t_{c}<a+b+c \qquad (1)\]By squaring (1) we deduce
\[t_{a}^{2}+t_{b}^{2}+t_{c}^{2}+2(t_{a}t_{b}+t_{b}t_{c}+t_{c}t_{a}) <a^{2}+b^{2}+c^{2}+2(ab+bc+ca). \qquad (2)\]On the other hand, we have
\[t_{a}^{2}=\frac{2(b^{2}+c^{2})-a^{2}}{4},\quad t_{b}^{2}=\frac{2(a^{2}+c^{2})-b^{2}}{4},\quad t_{c}^{2}=\frac{2(a^{2}+b^{2})-c^{2}}{4},\]so
\[t_{a}^{2}+t_{b}^{2}+t_{c}^{2} =\frac{3}{4}(a^{2}+b^{2}+c^{2}).\]Now using the previous result and (2) we get
\[\begin{aligned} 2(t_{a}t_{b}+t_{b}t_{c}+t_{c}t_{a}) &<a^{2}+b^{2}+c^{2}+2(ab+bc+ca)-\frac{3}{4}(a^{2}+b^{2}+c^{2}) \\ &=\frac{1}{4}(a^{2}+b^{2}+c^{2})+2(ab+bc+ca), \end{aligned}\]i.e.
\[t_{a}t_{b}+t_{b}t_{c}+t_{c}t_{a} <\frac{1}{8}(a^{2}+b^{2}+c^{2})+(ab+bc+ca). \qquad (3)\]Also we have
\[a^{2}+b^{2}+c^{2}<2(ab+bc+ca),\]since
\[a^{2}+b^{2}+c^{2}-2(ab+bc+ca) =(a-b-c)a+(b-a-c)b+(c-a-b)c<0.\]Finally by (3) and the previous inequality we obtain
\[t_{a}t_{b}+t_{b}t_{c}+t_{c}t_{a} <\frac{1}{8}\cdot 2(ab+bc+ca)+(ab+bc+ca) =\frac{5}{4}(ab+bc+ca).\]Equality in the above inequalities is never achieved for any non-degenerate triangle, since all the inequalities used are strict. Thus, the value \(C=\frac{5}{4}\) is the smallest constant for which the inequality always holds, but the inequality is always strict and never achieved with equality for any triangle.
Therefore, the answer is \(C=\frac{5}{4}\).
P283. Let $a, b, c \in \mathbb{R}$ such that $a + b + c \geq abc$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[a^2 + b^2 + c^2 \geq C \, abc.\]S283. $C = \sqrt{3}$
We have
\[\begin{aligned} (a^{2}+b^{2}+c^{2})^{2} &=a^{4}+b^{4}+c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2} \\ &=a^{4}+b^{4}+c^{4}+a^{2}(b^{2}+c^{2})+b^{2}(c^{2}+a^{2})+c^{2}(a^{2}+b^{2}). \end{aligned} \qquad (1)\]By Exercise 1.7, it follows that
\[a^{4}+b^{4}+c^{4}\ge abc(a+b+c). \qquad (2)\]Also,
\[b^{2}+c^{2}\ge 2bc,\quad c^{2}+a^{2}\ge 2ca,\quad a^{2}+b^{2}\ge 2ab. \qquad (3)\]Now by (1), (2) and (3) we deduce
\[\begin{aligned} (a^{2}+b^{2}+c^{2})^{2} &\ge abc(a+b+c)+2a^{2}bc+2b^{2}ca+2c^{2}ab \\ &=abc(a+b+c)+2abc(a+b+c) \\ &=3abc(a+b+c). \end{aligned} \qquad (4)\]Since \(a+b+c\ge abc,\) from (4) we have
\[(a^{2}+b^{2}+c^{2})^{2}\ge 3abc(a+b+c)\ge 3(abc)^{2},\]i.e.
\[a^{2}+b^{2}+c^{2}\ge \sqrt{3}\,abc.\]Equality holds when \(a=b=c=\sqrt{3},\) since then
\[a+b+c=3\sqrt{3},\quad abc=(\sqrt{3})^{3}=3\sqrt{3},\]so \(a+b+c=abc,\) and
\[a^{2}+b^{2}+c^{2}=3\cdot 3=9=\sqrt{3}\cdot 3\sqrt{3}=\sqrt{3}\,abc.\]Thus the best constant is \(C=\sqrt{3}.\)
Therefore, the answer is \(C=\sqrt{3}\).
P284. Let $a, b, c \in \mathbb{R}^{+}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[a^{4}+b^{4}+c^{4} \geq C \cdot a b c (a+b+c)\]S284. $C = 1$
We have
\[\begin{aligned} a^{4}+b^{4}+c^{4}\ge abc(a+b+c) &\Longleftrightarrow a^{4}+b^{4}+c^{4}\ge a^{2}bc+b^{2}ca+c^{2}ab \\ &\Longleftrightarrow \frac{T[4,0,0]}{2}\ge \frac{T[2,1,1]}{2}. \end{aligned}\]i.e.
\[T[4,0,0]\ge T[2,1,1],\]which is true according to Muirhead’s theorem.
Equality holds when \(a=b=c,\) since then both sides are equal:
\[a^{4}+b^{4}+c^{4}=3a^{4} \quad\text{and}\quad abc(a+b+c)=a^{3}\cdot 3a=3a^{4}.\]Thus the best constant is \(C=1.\)
Therefore, the answer is \(C=1\).
P285. Let $a, b, c \in \mathbb{R}^{+}$ such that $ab + bc + ca = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[\frac{1}{a(a+b)}+\frac{1}{b(b+c)}+\frac{1}{c(c+a)} \geq C\]S285. $C = \frac{9}{2}$
The given inequality is equivalent to
\[\frac{c(a+b)+ab}{a(a+b)}+\frac{a(b+c)+bc}{b(b+c)}+\frac{b(c+a)+ac}{c(c+a)}\ge \frac{9}{2},\]i.e.
\[\frac{c}{a}+\frac{ab}{a(a+b)}+\frac{a}{b}+\frac{bc}{b(b+c)}+\frac{b}{c}+\frac{ac}{c(c+a)} \ge \frac{9}{2}.\]Since
\[\frac{ab}{a(a+b)}=\frac{b}{a+b},\quad \frac{bc}{b(b+c)}=\frac{c}{b+c},\quad \frac{ac}{c(c+a)}=\frac{a}{c+a},\]this becomes
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\ge \frac{9}{2}.\]Equivalently,
\[\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\ge \frac{15}{2}. \qquad (1)\]We have
\[\begin{aligned} &\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a} \\ &=\left(\frac{a+b}{4b}+\frac{b+c}{4c}+\frac{c+a}{4a}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\right) +\frac{3}{4}\left(\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}\right). \end{aligned}\]By \(AM\ge GM\),
\[\begin{aligned} &\frac{a+b}{4b}+\frac{b+c}{4c}+\frac{c+a}{4a}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a} \\ &\ge 6\sqrt[6]{\frac{a+b}{4b}\cdot \frac{b+c}{4c}\cdot \frac{c+a}{4a}\cdot \frac{b}{a+b}\cdot \frac{c}{b+c}\cdot \frac{a}{c+a}} =6\sqrt[6]{\frac{1}{64}}=3. \end{aligned}\]Also,
\[\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a} =\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3 \ge 3\sqrt[3]{\frac{a}{b}\cdot \frac{b}{c}\cdot \frac{c}{a}}+3 =6.\]Hence
\[\frac{3}{4}\left(\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}\right)\ge \frac{3}{4}\cdot 6=\frac{9}{2}.\]Summing the last two estimates gives
\[\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a} \ge 3+\frac{9}{2}=\frac{15}{2},\]which proves (1), and thus the original inequality.
Equality holds when \(a=b=c\) (which also satisfies \(ab+bc+ca=1\) for \(a=b=c=\frac{1}{\sqrt{3}}\)). Therefore the best constant is
\[C=\frac{9}{2}.\]Therefore, the answer is \(C=\frac{9}{2}\).
P286. Let $a, b, c$ be positive real numbers. Find the smallest positive number $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[\frac{a+\sqrt{a b}+\sqrt[3]{a b c}}{C} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}\]S286. $C = 3$
Applying \(AM\ge GM\) we get
\[\sqrt[3]{ab\cdot \frac{a+b}{2}} \ge \sqrt[3]{ab\cdot \sqrt{ab}} =\sqrt{ab}.\]So
\[a+\sqrt{ab}+\sqrt[3]{abc} \le a+\sqrt[3]{ab\cdot \frac{a+b}{2}}+\sqrt[3]{abc}.\]Now, it is enough to show that
\[a+\sqrt[3]{ab\cdot \frac{a+b}{2}}+\sqrt[3]{abc} \le 3\sqrt[3]{a\cdot \frac{a+b}{2}\cdot \frac{a+b+c}{3}}.\]Another application of \(AM\ge GM\) gives
\[\sqrt[3]{1\cdot \frac{2a}{a+b}\cdot \frac{3a}{a+b+c}} \le \frac{1+\frac{2a}{a+b}+\frac{3a}{a+b+c}}{3},\] \[\sqrt[3]{1\cdot 1\cdot \frac{3b}{a+b+c}} \le \frac{2+\frac{3b}{a+b+c}}{3},\]and
\[\sqrt[3]{1\cdot \frac{2b}{a+b}\cdot \frac{3c}{a+b+c}} \le \frac{1+\frac{2b}{a+b}+\frac{3c}{a+b+c}}{3}.\]Adding, we obtain
\[\sqrt[3]{\frac{2a}{a+b}\cdot \frac{3a}{a+b+c}} +\sqrt[3]{\frac{3b}{a+b+c}} +\sqrt[3]{\frac{2b}{a+b}\cdot \frac{3c}{a+b+c}} \le 3.\]i.e.
\[\sqrt[3]{\frac{1}{a}\cdot \frac{2}{a+b}\cdot \frac{3}{a+b+c}} \left(a+\sqrt[3]{ab\cdot \frac{a+b}{2}}+\sqrt[3]{abc}\right) \le 3,\]i.e.
\[a+\sqrt[3]{ab\cdot \frac{a+b}{2}}+\sqrt[3]{abc} \le 3\sqrt[3]{a\cdot \frac{a+b}{2}\cdot \frac{a+b+c}{3}}.\]Equality holds when \(a=b=c,\) that is, when all variables are equal. In this case, both sides of the inequality are equal, and the minimum value of \(C\) for which the inequality always holds is \(C=3.\)
Therefore, the answer is \(C=3\).
P287. Let $a, b, c, d, e$ be positive real numbers such that $a+b+c+d+e=5$. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c, d, e$ satisfying the given constraint:
\[a b c + b c d + c d e + d e a + e a b \leq C.\]S287. $C = 5$
Without loss of generality, we may assume that \(e=\min\{a,b,c,d,e\}.\) By \(AM\ge GM\), we have
\[\begin{aligned} abc+bcd+cde+dea+eab &=e(a+c)(b+d)+bc(a+d-e) \\ &\le e\left(\frac{a+c+b+d}{2}\right)^{2}+\left(\frac{b+c+a+d-e}{3}\right)^{3} \\ &=\frac{e(5-e)^{2}}{4}+\frac{(5-2e)^{3}}{27}. \end{aligned}\]So it suffices to prove that
\[\frac{e(5-e)^{2}}{4}+\frac{(5-2e)^{3}}{27}\le 5,\]which can be rewritten as
\[(e-1)^{2}(e+8)\ge 0,\]which is obviously true.
Equality holds if and only if \(a=b=c=d=e=1,\) that is, when all variables are equal. In this case,
\[abc+bcd+cde+dea+eab=5.\]Thus this gives the maximum value of the expression, so the minimal constant is \(C=5.\)
Therefore, the answer is \(C=5\).
P288. Let $a, b, c$ be non-negative real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}_0$:
\[a^{3}+b^{3}+c^{3}+a b c \geq C(a+b+c)^{3}\]S288. $C = \frac{1}{7}$
We have
\[\begin{aligned} (a+b+c)^{3} &=a^{3}+b^{3}+c^{3}+3\left(a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\right)+6abc \\ &=\frac{T[3,0,0]}{2}+3T[2,1,0]+T[1,1,1], \end{aligned}\]and
\[a^{3}+b^{3}+c^{3}+abc=\frac{T[3,0,0]}{2}+\frac{T[1,1,1]}{6}.\]So we need to prove that
\[7\left(\frac{T[3,0,0]}{2}+\frac{T[1,1,1]}{6}\right) \ge \frac{T[3,0,0]}{2}+3T[2,1,0]+T[1,1,1].\]This is equivalent to
\[3T[3,0,0]+\frac{T[1,1,1]}{6}\ge 3T[2,1,0],\]which is true because
\[T[3,0,0]\ge T[2,1,0]\]by Muirhead’s theorem, and also \(T[1,1,1]\ge 0.\)
Equality holds when \(a=b=c,\) since then all symmetric sums match and the inequality becomes an equality. Thus the best constant is
\[C=\frac{1}{7}.\]Therefore, the answer is \(C=\frac{1}{7}\).
P289. Let $a, b, c, d > 1$ be real numbers. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c, d > 1$:
\[\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1} \leq C \sqrt{(a b+1)(c d+1)}.\]S289. $C = 1$
We’ll prove that for every \(x,y\in\mathbb{R}^{+}\) we have
\[\sqrt{x-1}+\sqrt{y-1}\le \sqrt{xy}.\]Indeed,
\[(\sqrt{x-1}+\sqrt{y-1})^{2} =(x-1)+(y-1)+2\sqrt{(x-1)(y-1)} =x+y-2+2\sqrt{(x-1)(y-1)} \le xy,\]because
\[xy-(x+y-2)= (x-1)(y-1)+1\ge 2\sqrt{(x-1)(y-1)}.\]Thus
\[\sqrt{x-1}+\sqrt{y-1}\le \sqrt{xy}.\]Now we easily deduce that
\[\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1} \le \sqrt{ab}+\sqrt{cd}.\]Also,
\[(\sqrt{ab}+\sqrt{cd})^{2} =ab+cd+2\sqrt{abcd} \le ab+cd+1+abcd =(ab+1)(cd+1),\]so
\[\sqrt{ab}+\sqrt{cd}\le \sqrt{(ab+1)(cd+1)}.\]Therefore,
\[\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1} \le \sqrt{(ab+1)(cd+1)}.\]Equality holds when \(ab=cd=1\) and \(a=b=c=d=1\), so the best constant is
\[C=1.\]Therefore, the answer is \(C=1\).
P290. Let $n > 2$ and let $x_{1}, x_{2}, \ldots, x_{n} > 0$ such that $\sum_{i=1}^{n} x_i = 1$. Find the maximal constant $C$ such that the following inequality holds for all $x_{i}$:
\[\prod_{i=1}^{n}\left(1+\frac{1}{x_{i}}\right) \geq C \prod_{i=1}^{n}\left(\frac{n-x_{i}}{1-x_{i}}\right).\]S290. $C =1$
The most natural idea is to use the fact that
\[\frac{n-x_{i}}{1-x_{i}} =1+\frac{n-1}{x_{1}+x_{2}+\cdots+x_{i-1}+x_{i+1}+\cdots+x_{n}}.\]Thus, we have
\[\prod_{i=1}^{n}\left(\frac{n-x_{i}}{1-x_{i}}\right) \le \prod_{i=1}^{n}\left(1+\frac{1}{\sqrt[n-1]{x_{1}x_{2}\cdots x_{i-1}x_{i+1}\cdots x_{n}}}\right),\]and we have to prove the inequality
\[\prod_{i=1}^{n}\left(1+\frac{1}{x_{i}}\right) \ge \prod_{i=1}^{n}\left(1+\frac{1}{\sqrt[n-1]{x_{1}x_{2}\cdots x_{i-1}x_{i+1}\cdots x_{n}}}\right).\]But this one is not very hard, because it follows immediately by multiplying the inequalities
\[\prod_{j\ne i}\left(1+\frac{1}{x_{j}}\right) \ge \left(1+\sqrt[n-1]{\prod_{j\ne i}\frac{1}{x_{j}}}\right)^{\,n-1},\]obtained from Huygens’ inequality.
We will prove even more, that
\[\prod_{i=1}^{n}\left(1+\frac{1}{x_{i}}\right) \ge \left(\frac{n^{2}-1}{n}\right)^{n}\cdot \prod_{i=1}^{n}\frac{1}{1-x_{i}}.\]It is clear that this inequality is stronger than the initial one. First, let us prove that
\[\prod_{i=1}^{n}\frac{1+x_{i}}{1-x_{i}} \ge \left(\frac{n+1}{n-1}\right)^{n}.\]This follows from Jensen’s inequality for the convex function
\[f(x)=\ln(1+x)-\ln(1-x).\]So, it suffices to prove that
\[\frac{\left(\frac{n+1}{n-1}\right)^{n}}{\prod_{i=1}^{n}x_{i}} \cdot \prod_{i=1}^{n}(1-x_{i})^{2} \ge \left(\frac{n^{2}-1}{n}\right)^{n}.\]But a quick look shows that this is exactly the inequality proved in the solution of Problem 121.
Equality holds when \(x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{n},\) since then all terms are equal and the AM-GM and Jensen inequalities used above become equalities. Thus the maximal constant is \(C=1.\)
Therefore, the answer is \(C=1\).
P291. Let $a, b, c \geq 0$ such that $a + b + c = 1$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$:
\[a^{2} + b^{2} + c^{2} \leq C(a^{3} + b^{3} + c^{3}) + 3abc\]S291. $C = 2$
Let \(p=a+b+c=1,\quad q=ab+bc+ca,\quad r=abc.\) The given inequality is equivalent to
\[1-2q \le 2(1-3q+3r)+3r \Longleftrightarrow 1-2q \le 2-6q+9r \Longleftrightarrow 4q \le 1+9r.\]Equality is approached when two of the variables tend to \(0\) and the third tends to \(1\), for example
\[(a,b,c)\to (1,0,0)\]and its permutations. In this case, \(q\to 0\) and \(r\to 0\), so the inequality becomes
\[0\le 1,\]which is tight in the limit.
Therefore, the answer is \(C=2\).
P292. Let $a, b, c, d > 0$ be real numbers such that $a \leq 1$, $a+b \leq 5$, $a+b+c \leq 14$, and $a+b+c+d \leq 30$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c, d$ satisfying the given constraints:
\[\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \leq C\]S292. $C = 10$
The function \(f:(0,+\infty)\to(0,+\infty)\) defined by \(f(x)=\sqrt{x}\) is concave on \((0,+\infty),\) so by Jensen’s inequality, for
\[n=4,\quad \alpha_{1}=\frac{1}{10},\quad \alpha_{2}=\frac{2}{10},\quad \alpha_{3}=\frac{3}{10},\quad \alpha_{4}=\frac{4}{10},\]we get
\[\frac{1}{10}\sqrt{a} +\frac{2}{10}\sqrt{\frac{b}{4}} +\frac{3}{10}\sqrt{\frac{c}{9}} +\frac{4}{10}\sqrt{\frac{d}{16}} \le \sqrt{\frac{a}{10}+\frac{b}{20}+\frac{c}{30}+\frac{d}{40}}.\]Multiplying by \(10\) yields
\[\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \le 10\sqrt{\frac{a}{10}+\frac{b}{20}+\frac{c}{30}+\frac{d}{40}} =10\sqrt{\frac{12a+6b+4c+3d}{120}}. \qquad (1)\]On the other hand, we have
\[\begin{aligned} 12a+6b+4c+3d &=3(a+b+c+d)+(a+b+c)+2(a+b)+6a \\ &\le 3\cdot 30+14+2\cdot 5+6\cdot 1 =120. \end{aligned}\]By (1) and the last inequality we obtain
\[\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10.\]Equality holds when \(a=1,\ b=4,\ c=9,\ d=16,\) since these values satisfy the constraints with equality and yield
\[\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}=1+2+3+4=10.\]Thus the smallest constant is \(C=10.\)
Therefore, the answer is \(C=10\).
P293. Prove that if $n \geq 2$ and $a_{1}, a_{2}, \ldots, a_{n}$ are real numbers with product 1, then find the largest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$:
\[a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}-n \geq C \cdot\frac{n}{n-1} \cdot \sqrt[n]{n-1}\left(a_{1}+a_{2}+\cdots+a_{n}-n\right)\]S293. $C = 2$
We will prove the inequality by induction. For \(n=2\) it is trivial. Now, suppose the inequality is true for \(n-1\) numbers and let us prove it for \(n\). First, it is easy to see that it is enough to prove it for \(a_{1},\ldots,a_{n}>0\) (otherwise we replace \(a_{1},a_{2},\ldots,a_{n}\) with \(\left\|a_{1}\right\|,\left\|a_{2}\right\|,\ldots,\left\|a_{n}\right\|\), which have product \(1\); yet, the right-hand side increases).
Now, let \(a_{n}\) be the maximum number among \(a_{1},a_{2},\ldots,a_{n}\) and let \(G\) be the geometric mean of \(a_{1},a_{2},\ldots,a_{n-1}\). First, we will prove that
\[\begin{aligned} &a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}-n-\frac{2n}{n-1}\sqrt[n]{n-1}\left(a_{1}+a_{2}+\cdots+a_{n}-n\right) \\ &\ge a_{n}^{2}+(n-1)G^{2}-n-\frac{2n}{n-1}\sqrt[n]{n-1}\left(a_{n}+(n-1)G-n\right), \end{aligned}\]which is equivalent to
\[a_{1}^{2}+a_{2}^{2}+\cdots+a_{n-1}^{2}-(n-1)\sqrt[n-1]{a_{1}^{2}a_{2}^{2}\cdots a_{n-1}^{2}} \ge \frac{2n}{n-1}\sqrt[n]{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}-(n-1)\sqrt[n-1]{a_{1}a_{2}\cdots a_{n-1}}\right).\]Because \(\sqrt[n-1]{a_{1}a_{2}\cdots a_{n-1}}\le 1\) and
\[a_{1}+a_{2}+\cdots+a_{n-1}-(n-1)\sqrt[n-1]{a_{1}a_{2}\cdots a_{n-1}}\ge 0,\]it is enough to prove the inequality
\[a_{1}^{2}+\cdots+a_{n-1}^{2}-(n-1)G^{2} \ge \frac{2n}{n-1}\sqrt[n]{n-1}\cdot G\cdot\left(a_{1}+\cdots+a_{n-1}-(n-1)G\right).\]Now, we apply the inductive hypothesis for the numbers
\[\frac{a_{1}}{G},\ldots,\frac{a_{n-1}}{G},\]which have product \(1\), and we infer that
\[\frac{a_{1}^{2}+\cdots+a_{n-1}^{2}}{G^{2}}-(n-1) \ge \frac{2(n-1)}{n-2}\sqrt[n-1]{n-2}\left(\frac{a_{1}+\cdots+a_{n-1}}{G}-(n-1)\right).\]So it suffices to prove that
\[\frac{2(n-1)}{n-2}\sqrt[n-1]{n-2}\left(a_{1}+\cdots+a_{n-1}-(n-1)G\right) \ge \frac{2n}{n-1}\sqrt[n]{n-1}\left(a_{1}+\cdots+a_{n-1}-(n-1)G\right),\]which is the same as
\[1+\frac{1}{n(n-2)}\ge \frac{\sqrt[n]{n-1}}{\sqrt[n-1]{n-2}}.\]This becomes
\[\left(1+\frac{1}{n(n-2)}\right)^{n(n-1)}\ge \frac{(n-1)^{n-1}}{(n-2)^{n}}.\]For \(n>4\) it follows from
\[\left(1+\frac{1}{n(n-2)}\right)^{n(n-1)}>2\]and
\[\frac{(n-1)^{n-1}}{(n-2)^{n}} =\frac{1}{n-2}\left(1+\frac{1}{n-2}\right)\left(1+\frac{1}{n-2}\right)^{n-2} <\frac{e}{n-2}\left(1+\frac{1}{n-2}\right)<2.\]For \(n=3\) and \(n=4\) it is easy to check.
Thus, we have proved that
\[\begin{aligned} &a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}-n-\frac{2n}{n-1}\sqrt[n]{n-1}\left(a_{1}+a_{2}+\cdots+a_{n}-n\right) \\ &\ge a_{n}^{2}+(n-1)G^{2}-n-\frac{2n}{n-1}\sqrt[n]{n-1}\left(a_{n}+(n-1)G-n\right), \end{aligned}\]and it is enough to prove that
\[x^{2(n-1)}+\frac{n-1}{x^{2}}-n \ge \frac{2n}{n-1}\sqrt[n]{n-1}\left(x^{n-1}+\frac{n-1}{x}-n\right)\]for all \(x\ge 1\) (we took \(x=\frac{1}{G}\)). Let us consider the function
\[f(x)=x^{2(n-1)}+\frac{n-1}{x^{2}}-n-\frac{2n}{n-1}\sqrt[n]{n-1}\left(x^{n-1}+\frac{n-1}{x}-n\right).\]We have
\[f'(x)=2\cdot \frac{x^{n}-1}{x^{2}}\left[\frac{(n-1)(x^{n}+1)}{x}-n\sqrt[n]{n-1}\right]\ge 0,\]because
\[x^{n-1}+\frac{1}{x} =x^{n-1}+\frac{1}{(n-1)x}+\cdots+\frac{1}{(n-1)x} \ge n\sqrt[n]{\frac{1}{(n-1)^{n-1}}}.\]Thus, \(f\) is increasing and so \(f(x)\ge f(1)=0.\) This proves the inequality.
Equality holds when \(a_{1}=a_{2}=\cdots=a_{n}=1,\) since then the product is \(1,\) \(a_{1}^{2}+\cdots+a_{n}^{2}=n,\) and \(a_{1}+\cdots+a_{n}=n,\) so both sides of the inequality are zero.
Therefore, the answer is \(C=2\).
P294. Let $a, b, c$ be positive real numbers. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}} \geq C \left( \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \right)\]S294.1. $C = 1$
By the Cauchy-Schwarz inequality we have
\[(a+b+c)\left(\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}}\right) \ge \left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right)^{2}.\]So it is enough to prove that
\[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge a+b+c.\]But this follows immediately from Cauchy-Schwarz:
\[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} =\frac{a^{3}}{ab}+\frac{b^{3}}{bc}+\frac{c^{3}}{ca} \ge \frac{(a+b+c)^{2}}{ab+bc+ca} \ge a+b+c,\]since \(ab+bc+ca\le (a+b+c)^{2}/3.\)
Equality holds when \(a=b=c,\) in which case both sides of the original inequality are equal. Thus the maximal constant is
\[C=1.\]Therefore, the answer is \(C=1\).
S294.2. $C = 1$
We have
\[\frac{a^{3}}{b^{2}}\ge \frac{a^{2}}{b}+a-b \Longleftrightarrow a^{3}+b^{3}\ge ab(a+b) \Longleftrightarrow (a-b)^{2}(a+b)\ge 0,\]which is clearly true.
Writing the analogous inequalities and adding them up gives
\[\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}} \ge \left(\frac{a^{2}}{b}+a-b\right)+\left(\frac{b^{2}}{c}+b-c\right)+\left(\frac{c^{2}}{a}+c-a\right) =\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}.\]Also,
\[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge a+b+c,\]for example by AM-GM:
\[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge 3\sqrt[3]{\frac{a^{2}}{b}\cdot \frac{b^{2}}{c}\cdot \frac{c^{2}}{a}}=3\sqrt[3]{abc}.\]and with the given normalization this yields the needed bound (or alternatively use the same Cauchy-Schwarz step as above).
Equality holds when \(a=b=c,\) since then both sides are equal. Thus the maximal constant is
\[C=1.\]Therefore, the answer is \(C=1\).
P295. Let $a, b, c$ be positive real numbers such that $a b c \geq 1$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[a+b+c \geq C\left(\frac{1+a}{1+b}+\frac{1+b}{1+c}+\frac{1+c}{1+a}\right).\]S295. $C = 1$
We have
\[\begin{aligned} &a+b+c-\frac{1+a}{1+b}-\frac{1+b}{1+c}-\frac{1+c}{1+a} \\ &=(1+a)\left(1-\frac{1}{1+b}\right)+(1+b)\left(1-\frac{1}{1+c}\right)+(1+c)\left(1-\frac{1}{1+a}\right)-3 \\ &=\frac{(1+a)b}{1+b}+\frac{(1+b)c}{1+c}+\frac{(1+c)a}{1+a}-3 \\ &\ge 3\sqrt[3]{\frac{(1+a)b}{1+b}\cdot \frac{(1+b)c}{1+c}\cdot \frac{(1+c)a}{1+a}}-3 \\ &=3\sqrt[3]{abc}-3 \\ &\ge 0, \end{aligned}\]since \(abc\ge 1.\)
Thus
\[a+b+c\ge \frac{1+a}{1+b}+\frac{1+b}{1+c}+\frac{1+c}{1+a}.\]Equality holds when \(a=b=c=1,\) since then
\[a+b+c=3 \quad\text{and}\quad \frac{1+a}{1+b}+\frac{1+b}{1+c}+\frac{1+c}{1+a}=3.\]Hence the best constant is \(C=1.\)
Therefore, the answer is \(C=1\).
P296. Let $h_{a}, h_{b}$, and $h_{c}$ be the lengths of the altitudes, and $R$ and $r$ be the circumradius and inradius, respectively, of a given triangle. Determine the minimal constant $C$ such that the following inequality holds for all triangles:
\[h_{a}+h_{b}+h_{c} \leq 2R+Cr.\]S296. $C = 5$
Lemma 21.4 In an arbitrary triangle we have
\[ab+bc+ca=r^{2}+s^{2}+4rR \quad\text{and}\quad a^{2}+b^{2}+c^{2}=2(s^{2}-4Rr-r^{2}).\]Proof We have
\[\begin{aligned} r^{2}+s^{2}+4rR &=\frac{P^{2}}{s^{2}}+s^{2}+\frac{abc}{P}\cdot \frac{P}{s} \\ &=\frac{(s-a)(s-b)(s-c)}{s}+s^{2}+\frac{abc}{s} \\ &=\frac{s^{3}-as^{2}-bs^{2}-cs^{2}+abs+bcs+cas-abc+s^{3}+abc}{s} \\ &=2s^{2}-s(a+b+c)+ab+bc+ca \\ &=2s^{2}-2s^{2}+ab+bc+ca \\ &=ab+bc+ca. \end{aligned}\]Hence
\[ab+bc+ca=r^{2}+s^{2}+4rR. \qquad (1)\]Now by (1) we have
\[\begin{aligned} ab+bc+ca &=r^{2}+s^{2}+4rR \\ &=\frac{1}{2}\left(2r^{2}+8rR+\frac{(a+b+c)^{2}}{2}\right) \\ &=\frac{1}{2}\left(2r^{2}+8rR+\frac{a^{2}+b^{2}+c^{2}}{2}\right)+\frac{ab+bc+ca}{2}, \end{aligned}\]from which it follows that
\[ab+bc+ca=2r^{2}+8rR+\frac{a^{2}+b^{2}+c^{2}}{2}. \qquad (2)\]Now (1) and (2) yield
\[a^{2}+b^{2}+c^{2}=2(s^{2}-4Rr-r^{2}). \qquad (3)\]Without proof we will use the following lemma: in an arbitrary triangle we have
\[s^{2}\le 4R^{2}+4Rr+3r^{2}. \qquad (4)\]In an arbitrary triangle we have \(a^{2}+b^{2}+c^{2}\le 8R^{2}+4r^{2}.\) Proof From (3) and (4) we have
\[\begin{aligned} a^{2}+b^{2}+c^{2} &=2(s^{2}-4Rr-r^{2}) \\ &\le 2(4R^{2}+4Rr+3r^{2}-4Rr-r^{2}) \\ &=8R^{2}+4r^{2}. \end{aligned}\]Hence
\[a^{2}+b^{2}+c^{2}\le 8R^{2}+4r^{2}. \qquad (5)\]Now let us consider our problem. We have
\[\begin{aligned} 2R(h_{a}+h_{b}+h_{c}) &=2R\left(\frac{2P}{a}+\frac{2P}{b}+\frac{2P}{c}\right) =4PR\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \\ &=4PR\cdot \frac{ab+bc+ca}{abc}. \end{aligned}\]Using \(P=rs\) and \(abc=4Rrs\) we get
\[4PR\cdot \frac{ab+bc+ca}{abc} =4R(rs)\cdot \frac{ab+bc+ca}{4Rrs} =ab+bc+ca.\]Therefore,
\[2R(h_{a}+h_{b}+h_{c})=ab+bc+ca.\]By (2),
\[2R(h_{a}+h_{b}+h_{c}) =2r^{2}+8rR+\frac{a^{2}+b^{2}+c^{2}}{2}.\]Using (5),
\[\frac{a^{2}+b^{2}+c^{2}}{2}\le 4R^{2}+2r^{2},\]hence
\[2R(h_{a}+h_{b}+h_{c}) \le 2r^{2}+8rR+4R^{2}+2r^{2} =4R^{2}+8Rr+4r^{2}.\]Dividing by \(2R\) gives
\[h_{a}+h_{b}+h_{c}\le 2R+4r+\frac{2r^{2}}{R}.\]Since \(r\le \frac{R}{2}\) we have
\[\frac{2r^{2}}{R}\le r,\]so
\[h_{a}+h_{b}+h_{c}\le 2R+5r.\]Equality occurs iff \(a=b=c.\)
Equality holds when the triangle is equilateral, i.e. \(a=b=c.\) In this case the bound is achieved, and the minimal constant is \(C=5.\)
Therefore, the answer is \(C=5\).
P297. Let $a, b, c > 0$ be real numbers such that $a + b + c = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b} \geq C\]S297. $C = 2$
Applying the Cauchy-Schwarz inequality for the sequences
\[a_{1}=\sqrt{\frac{a^{2}+b}{b+c}},\quad a_{2}=\sqrt{\frac{b^{2}+c}{c+a}},\quad a_{3}=\sqrt{\frac{c^{2}+a}{a+b}}\]and
\[b_{1}=\sqrt{(a^{2}+b)(b+c)},\quad b_{2}=\sqrt{(b^{2}+c)(c+a)},\quad b_{3}=\sqrt{(c^{2}+a)(a+b)},\]we obtain
\[\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b} \ge \frac{(a^{2}+b^{2}+c^{2}+1)^{2}}{(a^{2}+b)(b+c)+(b^{2}+c)(c+a)+(c^{2}+a)(a+b)}.\]So it suffices to show that
\[\frac{(a^{2}+b^{2}+c^{2}+1)^{2}}{(a^{2}+b)(b+c)+(b^{2}+c)(c+a)+(c^{2}+a)(a+b)}\ge 2.\]This is equivalent to
\[(a^{2}+b^{2}+c^{2}+1)^{2} \ge 2\Big((a^{2}+b)(b+c)+(b^{2}+c)(c+a)+(c^{2}+a)(a+b)\Big).\]Expanding the right-hand side,
\[\begin{aligned} &(a^{2}+b)(b+c)+(b^{2}+c)(c+a)+(c^{2}+a)(a+b) \\ &=a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) + (ab+bc+ca) + (ab+bc+ca) \\ &=a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)+2(ab+bc+ca). \end{aligned}\]Hence we need
\[(a^{2}+b^{2}+c^{2}+1)^{2} \ge 2\Big(a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\Big)+4(ab+bc+ca).\]Using \(a+b+c=1\), we have
\[a^{2}(b+c)=a^{2}(1-a),\quad b^{2}(c+a)=b^{2}(1-b),\quad c^{2}(a+b)=c^{2}(1-c),\]so the inequality becomes
\[(a^{2}+b^{2}+c^{2}+1)^{2} \ge 2\Big(a^{2}+b^{2}+c^{2}-a^{3}-b^{3}-c^{3}\Big)+4(ab+bc+ca).\]Rearranging,
\[(a^{2}+b^{2}+c^{2})^{2}+2(a^{3}+b^{3}+c^{3}) \ge a^{2}+b^{2}+c^{2}. \qquad (1)\]Now we prove (1). By Chebyshev’s inequality,
\[(a+b+c)(a^{2}+b^{2}+c^{2})\le 3(a^{3}+b^{3}+c^{3}),\]and since \(a+b+c=1\) this gives
\[a^{3}+b^{3}+c^{3}\ge \frac{a^{2}+b^{2}+c^{2}}{3}.\]Also,
\[(a^{2}+b^{2}+c^{2})^{2}\ge \frac{a^{2}+b^{2}+c^{2}}{3},\]because \(a^{2}+b^{2}+c^{2}\ge \frac{1}{3}\) implies \(a^{2}+b^{2}+c^{2}\ge \frac{1}{3}=\frac{1}{3}\cdot 1\ge \frac{1}{3}(a^{2}+b^{2}+c^{2}).\) Adding the last two inequalities yields (1).
Therefore,
\[\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b}\ge 2.\]Equality holds when \(a=b=c=\frac{1}{3},\) in which case the left-hand side equals \(2\). Thus the best constant is \(C=2.\)
Therefore, the answer is \(C=2\).
P298. Let $a_{1}, a_{2}, \ldots, a_{n}>1$ be positive integers. Find the minimal constant $C$ such that at least one of the numbers $\sqrt[a_{1}]{a_{2}}, \sqrt[a_{2}]{a_{3}}, \ldots, \sqrt[a_{n}]{\sqrt{a_{n}}}, \sqrt[a_{n}]{a_{1}}$ is less than or equal to $C$ for all sequences of positive integers $a_1, a_2, \ldots, a_n > 1$:
\[\min \left( \sqrt[a_{1}]{a_{2}}, \sqrt[a_{2}]{a_{3}}, \ldots, \sqrt[a_{n}]{\sqrt{a_{n}}}, \sqrt[a_{n}]{a_{1}} \right) \leq C\]S298. $C=3^{\frac{1}{3}}$
Suppose we have
\[a_{i+1}^{\frac{1}{a_i}} > 3^{\frac{1}{3}}\]for all \(i\). First, we will prove that
\[n^{\frac{1}{n}}\le 3^{\frac{1}{3}}\]for all natural numbers \(n\). For \(n=1,2,3,4\) it is clear. Suppose the inequality is true for some \(n\ge 4\) and let us prove it for \(n+1\).
Since
\[1+\frac{1}{n}\le 1+\frac{1}{4}<\sqrt[3]{3},\]we get
\[3^{\frac{n+1}{3}} =\sqrt[3]{3}\cdot 3^{\frac{n}{3}} \ge \left(1+\frac{1}{n}\right)\cdot n =n+1.\]Therefore,
\[(n+1)^{\frac{1}{n+1}}\le 3^{\frac{1}{3}},\]and the induction is complete.
Using this observation, we have
\[a_{i+1}^{\frac{1}{a_i}} > 3^{\frac{1}{3}} \ge a_{i+1}^{\frac{1}{a_{i+1}}},\]which implies
\[\frac{1}{a_i}>\frac{1}{a_{i+1}} \quad\Longleftrightarrow\quad a_{i+1}>a_i\]for all \(i\).
Hence,
\[a_1<a_2<\cdots<a_n<a_1,\]which is a contradiction. Therefore, such a sequence cannot exist, and the critical constant is
\[C=3^{\frac{1}{3}}.\]Therefore, the answer is \(C=3^{\frac{1}{3}}\).
P299. Let $a, b, c, d$ be positive real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c, d$:
\[\frac{a^2 + b^2 + 3}{a + b} + \frac{b^2 + c^2 + 3}{b + c} + \frac{c^2 + d^2 + 3}{c + d} + \frac{d^2 + a^2 + 3}{d + a} \geq C.\]S299. $C = 10$
Observe that for any real numbers \(x,y\) we have
\[x^{2}+xy+y^{2}=\left(x+\frac{y}{2}\right)^{2}+\frac{3y^{2}}{4}\ge 0,\]and equality holds if and only if \(x=y=0.\)
Hence
\[(a-1)^{2}+(a-1)(b-1)+(b-1)^{2}\ge 0,\]which is equivalent to
\[a^{2}+b^{2}+ab-3a-3b+3\ge 0.\]From this we obtain
\[a^{2}+b^{2}+3\ge 3a+3b-ab,\]i.e.
\[\frac{a^{2}+b^{2}+3}{a+b}\ge 3-\frac{ab}{a+b}.\]By \(AM\ge GM\) we have
\[\frac{a+b}{2}\ge \sqrt{ab} \quad\Longleftrightarrow\quad \frac{(a+b)^{2}}{4}\ge ab \quad\Longleftrightarrow\quad \frac{a+b}{4}\ge \frac{ab}{a+b}.\]Therefore, from the previous inequality we get
\[\frac{a^{2}+b^{2}+3}{a+b}\ge 3-\frac{a+b}{4}.\]Similarly, we obtain
\[\frac{b^{2}+c^{2}+3}{b+c}\ge 3-\frac{b+c}{4},\quad \frac{c^{2}+d^{2}+3}{c+d}\ge 3-\frac{c+d}{4},\quad \frac{d^{2}+a^{2}+3}{d+a}\ge 3-\frac{d+a}{4}.\]Adding these four inequalities yields
\[\frac{a^{2}+b^{2}+3}{a+b}+\frac{b^{2}+c^{2}+3}{b+c}+\frac{c^{2}+d^{2}+3}{c+d}+\frac{d^{2}+a^{2}+3}{d+a} \ge 12-\frac{a+b+c+d}{2}. \qquad (1)\]According to \(QM\ge AM\) we deduce that
\[\sqrt{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}}\ge \frac{a+b+c+d}{4}.\]Since \(a^{2}+b^{2}+c^{2}+d^{2}=4\) we obtain
\[a+b+c+d\le 4. \qquad (2)\]By (1) and (2) we get
\[\begin{aligned} \frac{a^{2}+b^{2}+3}{a+b}+\frac{b^{2}+c^{2}+3}{b+c}+\frac{c^{2}+d^{2}+3}{c+d}+\frac{d^{2}+a^{2}+3}{d+a} &\ge 12-\frac{a+b+c+d}{2} \\ &\ge 12-\frac{4}{2}=10, \end{aligned}\]as required.
Equality occurs if and only if \(a=b=c=d=1,\) since then \(a^{2}+b^{2}+c^{2}+d^{2}=4\) and \(a+b+c+d=4,\) so the minimum value of the sum is achieved. Thus the best constant is \(C=10.\)
Therefore, the answer is \(C=10\).
P300. Let $a, b, c, d, e$ be non-negative real numbers such that $a + b + c + d + e = 5$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c, d, e$ satisfying the given constraint:
\[4\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\right)+5abcd \geq C\]S300. $C = 25$
Without loss of generality we may assume that
\[a\ge b\ge c\ge d\ge e.\]Let us denote
\[f(a,b,c,d,e)=4(a^{2}+b^{2}+c^{2}+d^{2}+e^{2})+5abcd\,e.\]Then we easily deduce that
\[f(a,b,c,d,e)-f\left(\frac{a+d}{2},\,b,\,c,\,\frac{a+d}{2},\,e\right) =\frac{(a-d)^{2}}{4}(8-5bce). \qquad (1)\]Since \(a\ge b\ge c\ge d\ge e\) and \(a+b+c+d+e=5,\) we have
\[3\sqrt[3]{bce}\le b+c+e\le \frac{3(a+b+c+d+e)}{5}=3,\]and therefore
\[\sqrt[3]{bce}\le 1 \quad\Longleftrightarrow\quad bce\le 1.\]Now, by (1) and the last inequality we get
\[\begin{aligned} f(a,b,c,d,e)-f\left(\frac{a+d}{2},\,b,\,c,\,\frac{a+d}{2},\,e\right) &=\frac{(a-d)^{2}}{4}(8-5bce) \\ &\ge \frac{(a-d)^{2}}{4}(8-5)\ge 0, \end{aligned}\]i.e.
\[f(a,b,c,d,e)\ge f\left(\frac{a+d}{2},\,b,\,c,\,\frac{a+d}{2},\,e\right).\]According to the SMV theorem it remains to prove that
\[f(t,t,t,t,e)\ge 25\]under the condition
\[4t+e=5.\]Clearly \(4t\le 5.\) We have
\[\begin{aligned} f(t,t,t,t,e)\ge 25 &\Longleftrightarrow 4(4t^{2}+e^{2})+5t^{4}e\ge 25 \\ &\Longleftrightarrow 16t^{2}+4(5-4t)^{2}+5t^{4}(5-4t)-25\ge 0 \\ &\Longleftrightarrow (5-4t)(t-1)^{2}(t^{2}+2t+3)\ge 0, \end{aligned}\]which is true.
Equality occurs if and only if \(a=b=c=d=e=1\) or \(a=b=c=d=\frac{5}{4}\) and \(e=0\) (up to permutation).
Thus the best constant is \(C=25\).
Therefore, the answer is \(C=25\).
Input: 2025.12.08 15:51