# INEQUALITY'S BLOG

## August 21, 2011

### Problem 301 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 12:30 am

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)$$\displaystyle \geq \biggl(\frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}+\frac{y^{\frac{11}{4}}}{\sqrt[4]{z^4+x^4}}+\frac{z^{\frac{11}{4}}}{\sqrt[4]{x^4+y^4}}\biggl)^2$
Let $\displaystyle A=\frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}+\frac{y^{\frac{11}{4}}}{\sqrt[4]{z^4+x^4}}+\frac{z^{\frac{11}{4}}}{\sqrt[4]{x^4+y^4}}$
We have $\displaystyle \frac{x^{\frac{11}{4}}}{\sqrt[4]{y^4+z^4}}=\frac{(x^3)^{\frac{5}{4}}}{\sqrt[4]{x^4y^4+z^4x^4}}$
Therefore we get $\displaystyle A=\frac{(x^3)^{\frac{5}{4}}}{\sqrt[4]{x^4y^4+z^4x^4}}+\frac{(y^3)^{\frac{5}{4}}}{\sqrt[4]{y^4z^4+x^4y^4}}+\frac{(z^3)^{\frac{5}{4}}}{\sqrt[4]{z^4x^4+y^4z^4}}$
Because $\displaystyle \frac{5}{4}-\frac{1}{4}=1$ so from $Problem$ $Van Khea$ we get
$\displaystyle A\geq \frac{(x^3+y^3+z^3)^{\frac{5}{4}}}{\sqrt[4]{2(x^4y^4+y^4z^4+z^4x^4)}}$$\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 $x=y=z=1\Leftrightarrow a=b=c=1$