<s>[INST] Proving the inequality $\sum_{\text{cyc}} \frac{a}{a^2+b^3+c^3}\le\frac1{5abc}$ Let $a,b,c> 0$ be three real numbers such that $a+b+c=1$. I want to prove that 
$$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\le\frac{1}{5abc}.$$
My attempt: Using AM-GM on each denominator gives (here, LHS denotes the left-hand side)
$$LHS\le\frac{1}{3a^{\frac12}bc}+\frac{1}{3b^{\frac12}ac}+\frac{1}{3c^{\frac12}ab}.$$
However, I think that my attempt doesn't work because the last expression can get larger than $\frac{1}{5abc}$. In fact, if we multiply with $3abc$ then the original inequality is:
$$\sqrt a+\sqrt b+\sqrt c\le\frac35$$
However, as $a,b,c<1$, $$\sqrt a +\sqrt b + \sqrt c> a+b+c=1>\frac35.$$ So my upper bound is always larger than the given one.
 [/INST] Also, by AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a}{a^2+b^3+c^3}=\sum_{cyc}\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq\sum_{cyc}\frac{a}{3abc+a^2b+a^2c}=$$
$$=\sum_{cyc}\frac{1}{3bc+ab+ac}\leq\frac{1}{(3+1+1)^2}\sum_{cyc}\left(\frac{3^2}{3bc}+\frac{1^2}{ab}+\frac{1^2}{ac}\right)=\frac{1}{5}\sum_{cyc}\frac{1}{ab}=\frac{1}{5abc}.$$
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