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Let $$\displaystyle\left\{ \begin{array}{l} a,b,c>0\\ a+b+c=3 \end{array} \right.$$ Prove that: $$\frac{a^3}{(2a^2+b^2)(2a^2+c^2)}+\frac{b^3}{(2b^2+c^2)(2b^2+a^2)}+\frac{c^3}{(2c^2+a^2)(2c^2+b^2)}\leq\frac{1}{3}$$

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Notice first that: $$\frac{a^3}{(2a^2+b^2)(2a^2+c^2)}=\frac{1}{a}\frac{1}{(2+\frac{b^2}{a^2})(2+\frac{c^2}{a^2})}$$ Since,$$(2+\frac{b^2}{a^2})(2+\frac{c^2}{a^2})=(1+1+\frac{b^2}{a^2})(1+\frac{c^2}{a^2}+1)\ge(1+\frac{c}{a}+\frac{b}{a})^2=\frac{(a+b+c)^2}{a^2}$$ by Cauchy-Bunjakowski inequlity,we get the following: $$\frac{1}{(2+\frac{b^2}{a^2})(2+\frac{c^2}{a^2})}\le\frac{a^2}{(a+b+c)^2}$$ Maximizing each of the remaining two addents in the LHS of the given inequality in similar way we see that:

$$LHS \le \frac{1}{a}\frac{a^2}{(a+b+c)^2}+\frac{1}{b}\frac{b^2}{(a+b+c)^2}+\frac{1}{c}\frac{c^2}{(a+b+c)^2}=\frac{(a+b+c)}{(a+b+c)^2}=\frac{1}{3}$$

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