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

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