Let a, b, c be positive real numbers such that $a+b+c=1$. Prove that $\frac{a}{a^2 + b^3 + c^3} + \frac{b}{b^2 + c^3 + a^3} + \frac{c}{c^2 + a^3 + b^3} \le \frac{1}{5abc}$
This is a problem from RMO 2019. I tried using the method of Lagrange Multipliers, but it seems too lengthy. The solution on the site uses manipulative elementary proof.
Any non-elementary proofs or a straightforward method to solve these kind of problems?
Thank You.
Too long for comment:
Rewrite the inequality as: $$ 5\sum_{cyc}\frac{a^2bc}{a^2+b^3+c^3}\leq 1, $$ By AM-GM we have $$a^2bc \leq \frac{(1+a)^4}{256},$$ and by the power-means inequality we have $$ \left(\frac{b^3+c^3}{2}\right)^{1/3}\geq\frac{b+c}{2}\implies b^3+c^3\geq\frac{(1-a)^3}{4}, $$ so $$ 5\sum_{cyc}\frac{a^2bc}{a^2+b^3+c^3}\leq \frac{5}{64}\sum_{cyc}\frac{(1+a)^4}{4a^2+(1-a)^3} $$
The RHS seems to be $\leq 1$, but I don't know how to prove it. The function $f(t)=\frac{(1+t)^4}{4t^2+(1-t)^3}$ is almost concave. Perhaps it can be approximated by some "nice" concave function.