## Problem 303 Van Khea

If $a, b, x, y, z$ are positive real numbers such that $a+b=1$. Prove that
$\displaystyle x^3+y^3+z^3+6abxyz$$\geq (2a-b^2)(x^2y+y^2z+z^2x)+(2b-a^2)(xy^2+yz^2+zx^2)$

## Problem 302 Van Khea

If $a, b, c$ are positive real numbers such that $a+b+c=3$ then prove that
$29(a^2b+b^2c+c^2a)+38(ab^2+bc^2+ca^2)+42abc\leq 243$

## Problem 301 Van Khea

គេអោយ $a, b, c$ ជាចំនួនពិតវិជ្ជមាន។ ស្រាយបញ្ជាក់ថាៈ
Let $a, b, c$ be positive real numbers. Prove that:
$\displaystyle \frac{a^2}{b\sqrt{c^2+3ab}}+\frac{b^2}{c\sqrt{a^2+3bc}}+\frac{c^2}{a\sqrt{b^2+3ca}}\geq \frac{3}{2}$
Solution
We have
$\displaystyle \frac{a^2}{b\sqrt{c^2+3ab}}+\frac{b^2}{c\sqrt{a^2+3bc}}+\frac{c^2}{a\sqrt{b^2+3ca}}$$\displaystyle =\frac{(a^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ab)^3}}+\frac{(b^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(bc)^3}}+\frac{(c^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ca)^3}}$
Because $\displaystyle \frac{3}{2}-\frac{1}{2}=1$ then we have
$\displaystyle \frac{(a^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ab)^3}}+\frac{(b^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(bc)^3}}+\frac{(c^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ca)^3}}$$\displaystyle \geq \frac{(a^2+b^2+c^2)^{\frac{3}{2}}}{\sqrt{3((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)}}$
Thus, it suffices to show that
$\displaystyle \frac{(a^2+b^2+c^2)^{\frac{3}{2}}}{\sqrt{3((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)}}\geq \frac{3}{2}$
$\displaystyle \Leftrightarrow 27((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)\leq 4(a^2+b^2+c^2)^3$
Letting $a^2=x; b^2=y; c^2=z$ yields
$\displaystyle 27((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)\leq 4(x+y+z)^3$
From $Cauchy-Schwarz$ inequality we have
$\displaystyle ((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)^2$$\leq (x^2y+y^2z+z^2x+xyz)(xy^2+yz^2+zx^2+xyz)$
So we need to prove that
$\displaystyle (x^2y+y^2z+z^2x+xyz)(xy^2+yz^2+zx^2+xyz)\leq \frac{16}{27^2}(x+y+z)^6$
Now we will show that for $x, y, z$ are positive real numbers then we have
$27(x^2y+y^2z+z^2x+xyz)\leq 4(x+y+z)^3$ ;$(1)$ and $27(xy^2+yz^2+zx^2+xyz)\leq 4(x+y+z)^3$ ; $(2)$
without loss of generality, suppose that $x=min(x, y, z)$. Sitting $y=x+u$ and $z=x+v; (u, v\geq 0)$ then:
$(1)\Leftrightarrow 9(u^2-uv+v^2)x+(u-2v)^2(4u+v)\geq 0$
$(2)\Leftrightarrow 9(u^2-uv+v^2)x+(2u-v)^2(u+4v)\geq 0$
which is obviously true. Equality occurs for $u=v=0\Leftrightarrow x=y=z$
$\displaystyle \Rightarrow ((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)^2$$\displaystyle \leq \frac{16}{27^2}(x+y+z)^6$
$\displaystyle \Leftrightarrow 27((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)\leq 4(x+y+z)^3$ is true.
Therefore the proof is completed. Equality occurs for $x=y=z\Leftrightarrow a=b=c$

## Problem 300 Van Khea

If $a, b, c$ are positive real numbers such that $a^2+b^2+c^2+2abc=5$. Prove that
$\displaystyle \frac{1}{a^3+b^3+c^3}+\frac{1}{abc}\geq \frac{4}{3}$
Solution
We have
$\displaystyle \frac{1}{a^3+b^3+c^3}+\frac{1}{abc}=\frac{1}{a^3+b^3+c^3}+\frac{a^2+b^2+c^2+2abc}{5abc}$
$\displaystyle \Rightarrow \frac{1}{a^3+b^3+c^3}+\frac{a}{5bc}+\frac{b}{5ca}+\frac{c}{5ab}+\frac{2}{5}\geq \frac{4}{3}$
$\displaystyle \Leftrightarrow \frac{1}{a^3+b^3+c^3}+\frac{a}{5bc}+\frac{b}{5ca}+\frac{c}{5ab}\geq \frac{14}{15}$
Let $\displaystyle a=\sqrt[3]{\frac{3x}{x+y+z}}; b=\sqrt[3]{\frac{3y}{x+y+z}}; c=\sqrt[3]{\frac{3z}{x+y+z}}$ then we get $a^3+b^3+c^3=3$
Sitting $a, b, c$ yields
$\displaystyle \frac{1}{3}+\frac{\sqrt[3]{x+y+z}}{5\sqrt[3]{3}}\biggl(\sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\biggl)\geq \frac{14}{15}$
$\displaystyle \Leftrightarrow \frac{\sqrt[3]{x+y+z}}{\sqrt[3]{3}}\biggl(\sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\biggl)\geq 3$
From $AM-GM$ inequality we have
$\sqrt[3]{x+y+z}\geq \sqrt[3]{3\sqrt[3]{xyz}}$ and $\displaystyle \sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\geq 3\sqrt[3]{\frac{1}{\sqrt[3]{xyz}}}$
then we get
$\displaystyle \frac{\sqrt[3]{x+y+z}}{\sqrt[3]{3}}\biggl(\sqrt[3]{\frac{x}{yz}}+\sqrt[3]{\frac{y}{zx}}+\sqrt[3]{\frac{z}{xy}}\biggl)\geq \frac{\sqrt[3]{3\sqrt[3]{xyz}}}{\sqrt[3]{3}}.3\sqrt[3]{\frac{1}{\sqrt[3]{xyz}}}\geq 3$
Therefore the proof is completed. Equality occurs for $x=y=z\Leftrightarrow a=b=c=1$

## Problem 297 Van Khea

If $a, b, c$ are positive real numbers. Prove that:
$\displaystyle \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{1}{3}(a+b+c)^2$
Proof
From $Cauchy-Schwarz$ we have
$\displaystyle \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{(a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a})^2}{2(ab+bc+ca)}$
But for $a, b, c$ are positive real numbers then $3(ab+bc+ca)\leq (a+b+c)^2$
$\displaystyle \Rightarrow \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{3(a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a})^2}{2(a+b+c)^2}$
Thus, we need to prove that
$\displaystyle a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a}\geq \frac{\sqrt{2}}{3}(a+b+c)^2$
We have $\displaystyle a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a}$$\displaystyle =a^2\sqrt{1+\frac{b}{a}}+b^2\sqrt{1+\frac{c}{b}}+c^2\sqrt{1+\frac{a}{c}}$$\displaystyle =\frac{a^2}{\sqrt{\frac{a}{a+b}}}+\frac{b^2}{\sqrt{\frac{b}{b+c}}}+\frac{c^2}{\sqrt{\frac{c}{c+a}}}$$\displaystyle \geq \frac{(a+b+c)^2}{\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}}$
From $Problem$ $Vasile Cirtoaje$ we have
$\displaystyle \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\leq \frac{3}{\sqrt{2}}$
$\displaystyle \Rightarrow a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a}\geq \frac{\sqrt{2}}{3}(a+b+c)^2$
$\displaystyle \Rightarrow (a^{\frac{3}{2}}\sqrt{a+b}+b^{\frac{3}{2}}\sqrt{b+c}+c^{\frac{3}{2}}\sqrt{c+a})^2\geq \frac{2}{9}(a+b+c)^4$
$\displaystyle \Rightarrow \frac{a+b}{b+c}.a^2+\frac{b+c}{c+a}.b^2+\frac{c+a}{a+b}.c^2\geq \frac{1}{3}(a+b+c)^2$
Therefore the proof is completed. Equality occurs for $a=b=c$
របៀបទីពីរ ( Rainymathboy)
សន្មត $a\ge b\ge c$
ទាញបាន $\displaystyle {{a}^{2}}\ge {{b}^{2}}\ge {{c}^{2}}$
$\displaystyle a+b\ge c+a\ge b+c$ និង​ $\frac{1}{b+c}\ge \frac{1}{c+a}\ge \frac{1}{a+b}$
នោះ $\displaystyle \frac{{{a}^{2}}}{b+c}\ge \frac{{{b}^{2}}}{c+a}\ge \frac{{{c}^{2}}}{a+b}$
តាមវិសមភាព​តំរៀប (Rearrangement​ inequality) យើងមានៈ
បើ​ $\displaystyle {{a}_{1}}\ge {{a}_{2}}\ge \ldots \ge {{a}_{n}}$​ និង ${{b}_{1}}\ge {{b}_{2}}\ge \ldots \ge {{b}_{n}}$
តាង $\displaystyle A={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+\ldots +{{a}_{n}}{{b}_{n}}$
$\displaystyle B={{a}_{1}}{{b}_{\sigma (1)}}+{{a}_{2}}{{b}_{\sigma (2)}}+\ldots +{{a}_{n}}{{b}_{\sigma (n)}}$
$\displaystyle C={{a}_{1}}{{b}_{n}}+{{a}_{2}}{{b}_{n-1}}+\ldots +{{a}_{n}}{{b}_{1}}$
នាំអោយ $A\ge B\ge C$
យើងទាញបានៈ
$\displaystyle A=\frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(b+c)$
$\displaystyle B=\frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(b+c)+\frac{{{c}^{2}}}{a+b}(c+a)$
$\displaystyle C=\frac{{{a}^{2}}}{b+c}(b+c)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(a+b)$
យើងយកលក្ខខណ្ឌ​ $B\ge C$ មក​ប្រើ​នោះយើងទាញបានៈ
$\displaystyle \frac{{{a}^{2}}}{b+c}(a+b)+\frac{{{b}^{2}}}{c+a}(b+c)+\frac{{{c}^{2}}}{a+b}(c+a)$
$\displaystyle \ge \frac{{{a}^{2}}}{b+c}(b+c)+\frac{{{b}^{2}}}{c+a}(c+a)+\frac{{{c}^{2}}}{a+b}(a+b)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$
$\displaystyle a^2+b^2+c^2\geq \frac{1}{3}(a+b+c)^2$
ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។ សមភាពកើតមានពេល $a=b=c$

## Problem 295 Van Khea (Special problem)

If $a, b, c$ are positive real numbers such that $abc=1$. Prove that
$\displaystyle \frac{a^{\frac{5}{8}}}{\sqrt{b+c}}+\frac{b^{\frac{5}{8}}}{\sqrt{c+a}}+\frac{c^{\frac{5}{8}}}{\sqrt{a+b}}\geq \frac{3}{\sqrt{2}}$
Proof
Let $a=x^4; b=y^4; c=z^4\Rightarrow xyz=1$ then we get:
$\displaystyle \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{3}{\sqrt{2}}$
From $Cauchy-Scharz$ inequality we have
$\displaystyle (x^3+y^3+z^3)\biggl(\frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\biggl)$$\displaystyle \geq \biggl(\frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}+\frac{y^{\frac{11}{4}}}{\sqrt[4]{z^4+x^4}}+\frac{z^{\frac{11}{4}}}{\sqrt[4]{x^4+y^4}}\biggl)^2$
Let $\displaystyle A=\frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}+\frac{y^{\frac{11}{4}}}{\sqrt[4]{z^4+x^4}}+\frac{z^{\frac{11}{4}}}{\sqrt[4]{x^4+y^4}}$
We have $\displaystyle \frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}=\frac{(x^3)^{\frac{5}{4}}}{\sqrt[4]{x^4y^4+z^4x^4}}$
Therefore we get $\displaystyle A=\frac{(x^3)^{\frac{5}{4}}}{\sqrt[4]{x^4y^4+z^4x^4}}+\frac{(y^3)^{\frac{5}{4}}}{\sqrt[4]{y^4z^4+x^4y^4}}+\frac{(z^3)^{\frac{5}{4}}}{\sqrt[4]{z^4x^4+y^4z^4}}$
Because $\displaystyle \frac{5}{4}-\frac{1}{4}=1$ so from $Problem$ $Van Khea$ inequality we get
$\displaystyle A\geq \frac{(x^3+y^3+z^3)^{\frac{5}{4}}}{\sqrt[4]{2(x^4y^4+y^4z^4+z^4x^4)}}$$\displaystyle \Rightarrow A^2\geq \frac{(x^3+y^3+z^3)^{\frac{5}{2}}}{\sqrt{2(x^4y^4+y^4z^4+z^4x^4)}}$
$\displaystyle \Rightarrow (x^3+y^3+z^3)\biggl(\frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\biggl)$$\displaystyle \geq \frac{(x^3+y^3+z^3)^{\frac{5}{2}}}{\sqrt{2(x^4y^4+y^4z^4+z^4x^4)}}$
$\displaystyle \Rightarrow \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{(x^3+y^3+z^3)^{\frac{3}{2}}}{\sqrt{2(x^4y^4+y^4z^4+z^4x^4)}}$
Now we will prove that if $x, y, z$ be positive real numbers then
$\displaystyle 3^{\frac{5}{3}}(x^4y^4+y^4z^4+z^4x^4)\leq (x^3+y^3+z^3)^{\frac{8}{3}}$
Let $\displaystyle u=\frac{x\sqrt[3]{3}}{\sqrt[3]{x^3+y^3+z^3}};v=\frac{y\sqrt[3]{3}}{\sqrt[3]{x^3+y^3+z^3}};w=\frac{z\sqrt[3]{3}}{\sqrt[3]{x^3+y^3+z^3}}$ then we just need to prove that with $u^3+v^3+w^3=3$ then $(uv)^4+(vw)^4+(wu)^4\leq 3$
From $AM-GM$ inequality we have
$\displaystyle uv\leq \frac{u^3+v^3+1}{3}=\frac{4-w^3}{3}$$\displaystyle \Rightarrow (uv)^4\leq \frac{4u^3v^3-u^3v^3w^3}{3}$
Therefore we get $\displaystyle (uv)^4+(vw)^4+(wu)^4\leq \frac{4((uv)^3+(vw)^3+(wu)^3)}{3}-u^3w^3w^3$
Thus, it suffices to show that: $4((uv)^3+(vw)^3+(wu)^3)-3u^3v^3w^3\leq 9$
Which is just the third degree $Schur's$ inequality
$4(rs+st+tr)(r+s+t)-3rst\leq (r+s+t)^3$
For $r=u^3; s=v^3; t=w^3$
Therefore we get $\displaystyle 3^{\frac{5}{3}}(x^4y^4+y^4z^4+z^4x^4)\leq (x^3+y^3+z^3)^{\frac{8}{3}}$
$\displaystyle \Rightarrow \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{3^{\frac{5}{6}}(x^3+y^3+z^3)^{\frac{3}{2}}}{(x^3+y^3+z^3)^{\frac{4}{3}}\sqrt{2}}$$\displaystyle \geq \frac{3^{\frac{5}{6}}(x^3+y^3+z^3)^{\frac{1}{6}}}{\sqrt{2}}$
From $AM-GM$ inequality we have $x^3+y^3+z^3\geq 3xyz=3$
$\displaystyle \Rightarrow \frac{x^{\frac{5}{2}}}{\sqrt{y^4+z^4}}+\frac{y^{\frac{5}{2}}}{\sqrt{z^4+x^4}}+\frac{z^{\frac{5}{2}}}{\sqrt{x^4+y^4}}\geq \frac{3}{\sqrt{2}}$
Therefore the proof is completed. Equality occurs for $x=y=z=1\Leftrightarrow a=b=c=1$