INEQUALITY'S BLOG

August 23, 2011

Problem 304 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 12:52 pm

If a, b, c are positive real numbers such that a^2+b^2+c^2+2abc=5 then 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

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