Inequality Proof Problems [251-300]

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Restructured the IneqMath training data.

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

\[a^{4}+b^{4}+c^{4}+d^{4}+C a b c d \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}+a^{2} c^{2}+b^{2} d^{2}\]

(Difficulty: Hard)

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 $\triangle ABC$:

\[\frac{L P}{R^3} \leq C.\]

(Difficulty: Medium)

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:

\[\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+d)}+\frac{1}{d(1+a)} \geq C\]

(Difficulty: Medium)

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 $x, y, z \in \mathbb{R}$:

\[x^{4}+y^{4}+z^{4} \geq 4 x y z + C\]

(Difficulty: Easy)

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:

\[\frac{1+a^{2}}{1-a^{2}}+\frac{1+b^{2}}{1-b^{2}}+\frac{1+c^{2}}{1-c^{2}} \geq C\]

(Difficulty: Hard)

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 $a, b, c$:

\[\frac{abc}{(1+a)(a+b)(b+c)(c+16)} \leq C.\]

(Difficulty: Easy)

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 $a, b, c \in (1,2)$:

\[\frac{b \sqrt{a}}{4 b \sqrt{c}-c \sqrt{a}}+\frac{c \sqrt{b}}{4 c \sqrt{a}-a \sqrt{b}}+\frac{a \sqrt{c}}{4 a \sqrt{b}-b \sqrt{c}} \geq C\]

(Difficulty: Medium)

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 $a, b, c$:

\[\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geq C\]

(Difficulty: Hard)

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:

\[2 x^{4} + C \geq 2 x^{3} + x^{2}\]

(Difficulty: Easy)

S259-1. $C = 1$

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 $C=1$.

S259-2.

To find the smallest constant $C$ such that the inequality

\[2x^4 + C \geq 2x^3 + x^2\]

holds for all real numbers $x$, we define the function

\[f(x) = 2x^4 + C - 2x^3 - x^2.\]

We need $f(x) \geq 0$ for all $x$.

First, find the critical points of $f(x)$ by taking the derivative:

\[f'(x) = \frac{d}{dx}(2x^4 - 2x^3 - x^2) = 8x^3 - 6x^2 - 2x.\]

Set $f’(x) = 0$:

\[8x^3 - 6x^2 - 2x = 0.\]

Factor out $2x$:

\[2x(4x^2 - 3x - 1) = 0.\]

This gives $x = 0$ or solving $4x^2 - 3x - 1 = 0$ using the quadratic formula:

\[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} = \frac{3 \pm \sqrt{9 + 16}}{8} = \frac{3 \pm 5}{8}.\]

Thus, the roots are:

\[x = 1 \quad \text{and} \quad x = -\frac{1}{4}.\]

Evaluate $f(x)$ at these critical points to find the minimum value of $C$:

At $x = 0$:

\[f(0) = 2(0)^4 + C - 2(0)^3 - (0)^2 = C.\]

At $x = 1$:

\[f(1) = 2(1)^4 + C - 2(1)^3 - (1)^2 = 2 + C - 2 - 1 = C - 1.\]

At $x = -\frac{1}{4}$:

\[f\left(-\frac{1}{4}\right) = 2\left(-\frac{1}{4}\right)^4 + C - 2\left(-\frac{1}{4}\right)^3 - \left(-\frac{1}{4}\right)^2 = C - \frac{3}{128}.\]

For $f(x)$ to be non-negative for all $x$, the smallest value of $C$ must satisfy:

\[C \geq 1 \quad \text{and} \quad C \geq \frac{3}{128}.\]

The larger of these is $C \geq 1$. Therefore, the smallest constant $C$ such that the inequality holds for all $x$ is:

\[\boxed{1}\]

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 $a, b, c$:

\[\frac{a^{3}+2}{b+2}+\frac{b^{3}+2}{c+2}+\frac{c^{3}+2}{a+2} \geq C.\]

(Difficulty: Hard)

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:

\[\frac{l_{\alpha}}{a}+\frac{l_{\beta}}{b}+\frac{l_{\gamma}}{c} \leq C\frac{s}{r}.\]

(Difficulty: Hard)

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:

\[a^{2}+b^{2}+c^{2} < C(a b+b c+c a)\]

(Difficulty: Easy)

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:

\[\sqrt[3]{a b + b c} + \sqrt[3]{b c + c a} + \sqrt[3]{c a + a b} \leq C\]

(Difficulty: Medium)

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:

\[\sqrt[n]{x_{1} x_{2} \cdots x_{n}} \geq C\]

(Difficulty: Medium)

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:

\[\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq C\]

(Difficulty: Medium)

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:

\[\left(\frac{3-x}{1-x}\right)^{2}+\left(\frac{3-y}{1-y}\right)^{2}+\left(\frac{3-z}{1-z}\right)^{2} > C\]

(Difficulty: Medium)

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 $a, b \in \mathbb{R}^{+}$:

\[C(a^3 + b^3) \geq (a + b)^3\]

(Difficulty: Easy)

S267-1. $C = 4$

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 $C=4$.

S267-2.

To solve the problem, we need to find the smallest constant $C$ such that the inequality

\[C(a^3 + b^3) \geq (a + b)^3\]

holds for all positive real numbers $a$ and $b$.

First, expand the right-hand side:

\[(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3.\]

We want this to be less than or equal to the left-hand side:

\[C(a^3 + b^3).\]

Thus, the inequality becomes:

\[C(a^3 + b^3) \geq a^3 + 3a^2b + 3ab^2 + b^3.\]

Rearranging gives:

\[(C-1)(a^3 + b^3) \geq 3a^2b + 3ab^2.\]

Factor the right-hand side:

\[3a^2b + 3ab^2 = 3ab(a + b).\]

The inequality now is:

\[(C-1)(a^3 + b^3) \geq 3ab(a + b).\]

To find the smallest $C$, consider the case where $a = b$. If $a = b = x$, the inequality becomes:

\[(C-1)(2x^3) \geq 6x^3.\]

Dividing both sides by $2x^3$ (which is positive since $x > 0$), we obtain:

\[C-1 \geq 3.\]

Thus, $C \geq 4$.

To verify that $C = 4$ is sufficient, substitute $C = 4$ into the inequality:

\[4(a^3 + b^3) \geq a^3 + b^3 + 3a^2b + 3ab^2.\]

This simplifies to:

\[3(a^3 + b^3) \geq 3a^2b + 3ab^2.\]

Dividing through by 3, we have:

\[a^3 + b^3 \geq a^2b + ab^2.\]

This inequality holds by the AM-GM inequality, since:

\[a^3 + b^3 \geq 2a^{3/2}b^{3/2} \geq a^2b + ab^2\]

for all $a, b > 0$. Therefore, the smallest constant $C$ such that the inequality holds for all $a, b > 0$ is $\boxed{4}$.

P268. Let $x, y, z > 0$ be real numbers. Determine the maximal constant $C$ such that the following inequality holds for all $x, y, z$:

\[(x y + y z + z x) \left( \frac{1}{(x + y)^2} + \frac{1}{(y + z)^2} + \frac{1}{(z + x)^2} \right) \geq C.\]

(Difficulty: Hard)

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 $a, b, c$:

\[(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c} \leq C (a^{a} b^{b} c^{c}).\]

(Difficulty: Easy)

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 $x, y, z$:

\[\frac{1}{x^{n}} + \frac{1}{y^{n}} + \frac{1}{z^{n}} \geq C (x^{n} + y^{n} + z^{n}).\]

(Difficulty: Hard)

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 $x, y, z$:

\[\frac{1}{1-x y}+\frac{1}{1-y z}+\frac{1}{1-z x} \leq C.\]

(Difficulty: Hard)

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 $\alpha_{i}$:

\[\alpha_{1}^{\alpha_{1}} \alpha_{2}^{\alpha_{2}} \cdots \alpha_{n}^{\alpha_{n}} \geq C.\]

(Difficulty: Medium)

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

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

S272-2. Jensen Inequality

To solve this problem, we need to find the maximal constant $C$ such that the inequality

\[\alpha_1^{\alpha_1} \alpha_2^{\alpha_2} \cdots \alpha_n^{\alpha_n} \geq C\]

holds for all positive real numbers $\alpha_i$ satisfying $\alpha_1 + \alpha_2 + \cdots + \alpha_n = 1$.

Let’s first consider the symmetric case where all $\alpha_i$ are equal. Since $\alpha_1 + \alpha_2 + \cdots + \alpha_n = 1$, if all $\alpha_i$ are equal, then $\alpha_i = \frac{1}{n}$ for each $i = 1, 2, \ldots, n$.

In this case, the expression becomes:

\[\left(\frac{1}{n}\right)^{\frac{1}{n}} \left(\frac{1}{n}\right)^{\frac{1}{n}} \cdots \left(\frac{1}{n}\right)^{\frac{1}{n}} = \left(\frac{1}{n}\right)^{\frac{1}{n} \cdot n} = \frac{1}{n}.\]

Thus, for this symmetric case, the expression evaluates to $\frac{1}{n}$.

Next, we consider deviations from this symmetric case. By the method of Lagrange multipliers or by Jensen’s inequality (since the function $f(x) = x \log x$ is convex for $x > 0$), we can argue that the product $\alpha_1^{\alpha_1} \alpha_2^{\alpha_2} \cdots \alpha_n^{\alpha_n}$ is minimized when all $\alpha_i$ are equal, i.e., when $\alpha_i = \frac{1}{n}$ for all $i$.

Since the symmetric case $\alpha_i = \frac{1}{n}$ for all $i$ gives the minimum value of the expression, and this value is $\frac{1}{n}$, the maximal constant $C$ for which the inequality holds for all $\alpha_i$ is indeed $\frac{1}{n}$.

Therefore, the maximal constant $C$ is

\[\boxed{\frac{1}{n}}.\]

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 $n \in \mathbb{N}$:

\[(a+b)^{n}\left(a^{n}+b^{n}\right) \leq C\left(a^{2 n}+b^{2 n}\right)\]

(Difficulty: Medium)

S273-1. $C = 2^n$

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 $C=2^{n}$.

S273-2.

It is a typical rearrangement inequality problem.

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 $a_{1}, a_{2}, \ldots, a_{n}$:

\[\sum_{k=1}^{n} k a_{k} \leq C + \sum_{k=1}^{n} a_{k}^{k}\]

(Difficulty: Easy)

S274-1. $C = \binom{n}{2}$

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

S274-2. Creative solution

To solve the problem, we need to find the smallest constant $C$ such that the inequality

\[\sum_{k=1}^{n} k a_{k} \leq C + \sum_{k=1}^{n} a_{k}^{k}\]

holds for all positive real numbers $a_1, a_2, \ldots, a_n$.

Introduce the new variables $b_k = a_k^k$. Then, we have $a_k = b_k^{1/k}$. Substitute these into the inequality:

\[\sum_{k=1}^{n} k a_{k} = \sum_{k=1}^{n} k b_k^{1/k}\]

Thus, the inequality becomes:

\[\sum_{k=1}^{n} k b_k^{1/k} \leq C + \sum_{k=1}^{n} b_k\]

For each term $k b_k^{1/k}$, consider the function $f(x) = k x^{1/k}$. We need to compare $f(x)$ with $x$ to understand how $f(x)$ behaves relative to $x$.

Consider the function $g(x) = k x^{1/k} - x$. We want to find the maximum value of $g(x)$ for $x > 0$.

Calculate the derivative:

\[g'(x) = \frac{1}{k} x^{1/k - 1} - 1\]

Setting ( g’(x) = 0 ) for finding critical points leads to $x = 1$. We have $g’(x) > 0$ for $x < 1$ (function increasing) and $g’(x) < 0$ for $x > 1$ (function decreasing)

Thus, $x = 1$ is a maximum point for $g(x)$, and the maximum value is $k - 1$.

The maximum value of $g(x) = k x^{1/k} - x$ is $k - 1$. Therefore, for each $k$, the inequality

\[k b_k^{1/k} \leq (k - 1) + b_k\]

holds. Summing over all $k$, we get:

\[\sum_{k=1}^{n} k b_k^{1/k} \leq \sum_{k=1}^{n} (k - 1) + \sum_{k=1}^{n} b_k\]

Thus, the smallest constant $C$ that satisfies the original inequality is:

\[C = \sum_{k=1}^{n} (k - 1) = \frac{n(n-1)}{2}\]

Therefore, the smallest constant $C$ such that the inequality holds for all positive real numbers $a_1, a_2, \ldots, a_n$ is

\[\boxed{\frac{n(n-1)}{2}}\]

P275. Let $ABCD$ be a cyclic quadrilateral. Determine the smallest constant $C$ such that the following inequality holds if and only if $AB \cdot BC = 2AD \cdot DC$:

\[BD^2 \leq C \cdot AC^2\]

P276. Hanson-Wright inequality

P277. Poincaré inequality