Maximum value of $a^2c^3+b^2d^3$. Let $a,b,c,d$ be non negative real numbers such that $a^5 + b^5 \le 1$ and $c^5 + d^5 \le 1$. Find the maximum possible value of $a^2c^3 + b^2d^3$.
I tried using AM, GM and some other basic inequalities but they were of no use. Need some hints.

It can be done easily with Holder's inequality, with $p=5/2, q=5/3$ and $u=(a^2,b^2), v=(c^3,d^3)$ then:
$$\begin{align}a^2c^3+b^2d^3 &= |(a^2c^3,b^2d^3)|_1\\&\leq |(a^2,b^2)|_p |(c^3,d^3)|_q \\&= (a^5+b^5)^{2/5}(c^5+d^5)^{3/5}\leq 1
\end{align}$$
It now suffices to find a case where it is equal to $1$, which is pretty easy to do.

A more elementary approach is to use AM/GM and get that:
$$\frac{2}{5}a^5+\frac{3}{5}c^5= \frac{1}{5}(a^5+a^5+c^5+c^5+c^5)\geq \sqrt[5]{a^{10}c^{15}} = a^2c^3$$
Similarly:
$$\frac{2}{5}b^5+\frac{3}{5}d^5\geq  b^2d^3$$
Adding you get:
$$1 = \frac{2}{5}+\frac{3}{5}\geq \frac{2}{5}(a^5+b^5)+\frac{3}{5}(c^5+d^5)\geq a^2c^3+b^2d^3$$
