<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] Multiply both sides with $abc$ to get that your inequality is equivalent to
\begin{equation}\tag 1\label 1\sum_{\text{cyc}} \frac{a^2bc}{a^2+b^3+c^3}\le\frac15.\end{equation}
Now note that, since $a+b+c=1$, we have $$a^2+b^3+c^3=a^2\cdot(a+b+c)+b^3+c^3=a^3+b^3+c^3+a^2b+a^2c.$$
Hence, by AM-GM inequality for 5 variables,
$$a^2+b^3+c^3\geq5\sqrt[5]{a^7b^4c^4}.$$
Hence, \begin{split}\sum_{\text{cyc}} \frac{a^2bc}{a^2+b^3+c^3}&\le\frac15\sum_{\text{cyc}} \frac{a^2bc}{\sqrt[5]{a^7b^4c^4}}\\&=\frac15\sum_{\text{cyc}} \sqrt[5]{a^3bc}\\&\overset{\text{AM-GM}}\le\frac1{25}\sum_{\text{cyc}}3a+b+c\\&=\frac{5(a+b+c)}{25}\\&\overset{a+b+c=1}=\frac15. \end{split}
Done!
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