<s>[INST] Prove the given inequality if $a+b+c=1$ Let $a,b,c$ be positive real numbers such that $a+b+c=1$, then prove that
$\frac{a}{a^2 +b^3 +c^3}+\frac{b}{a^3 +b^2 +c^3}+\frac{c}{a^3 +b^3 +c^2} \leq \frac{1}{5abc}$
Please provide some hint to proceed. I have used Arithmetic Mean-Geometric Mean inequality to proceed in such questions, but I am not getting how to use it if it is to be used. Kindly provide some directon.
Usually the questions I have dealt so far were like if $a+b+c=1$, find maximum value of $a^3b^4c^2$ which I find using weighted A.M-G.M, but here I am not able to gather the thoughts to proceed 
 [/INST] We need to prove that:
$$\sum_{cyc}\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq\frac{a+b+c}{5abc}$$ or
$$\sum_{cyc}\left(\frac{a}{5abc}-\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\right)\geq0$$ or
$$\sum_{cyc}\frac{a(a^3+b^3+c^3+a^2b+a^2c-5abc)}{a^3+b^3+c^3+a^2b+a^2c}\geq0$$ or
$$\sum_{cyc}\frac{a((a-b)(a^2+5ac-2ab-2b^2)-(c-a)(a^2+5ab-2ac-2c^2))}{a^3+b^3+c^3+a^2b+a^2c}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a^3+5a^2c-2a^2b-2b^2a}{a^3+b^3+c^3+a^2b+a^2c}-\frac{ b^3+5b^2c-2b^2a-2a^2b}{a^3+b^3+c^3+b^2c+b^2a}\right)\geq0$$ or
$$\sum_{cyc}\tfrac{(a-b)^2(a^5+a^4b+5a^4b^2+4a^2b^3+ab^4+b^5+(5a^4+7a^3b+10a^2b^2+7ab^3+5b^4)c+(a^2+ab+b^2)c^3+5(a+b)c^4)}{(a^3+b^3+c^3+a^2b+a^2c)(a^3+b^3+c^3+b^2c+b^2a)}\geq0.$$
Done!
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