Prove: $\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+c^3+a^3}+\frac{c}{c^2+a^3+b^3}\leq \frac{1}{5abc}$ for $a+b+c=1$.

Question

Source: RMO 2019, question 3

Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+c^3+a^3}+\frac{c}{c^2+a^3+b^3}\leq \frac{1}{5abc}.$$

I tried using Holder's inequality but couldn't get to the desired result.

Answer 1

Hint: Use AM-GM to prove the following ineq: for all $a,b,c>0$, $$\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq \frac{3a+b+c}{25abc}.$$