TITLE: Prove the inequality using AM-GM inequality
QUESTION [6 upvotes]: Given that $a,b,u,v \geq 0$ and $$a^5+b^5 \leq 1$$ $$u^5+v^5 \leq 1$$ Prove that $$a^2u^3+b^2v^3 \leq 1$$ This looks like Holder's inequality, but I found this problem in a book just after the AM-GM section. So, I wondered whether it could be solved using AM-GM (I wasn't able to do it, I don't know why- I am unable to get the square and the cubic term). Thanks.

REPLY [7 votes]: Probably the most straightforward approach is given by Young's inequality with $p=\frac{5}{3},q=\frac{5}{2}$:
$$ a^2 u^3 \leq \frac{2}{5}a^5 + \frac{3}{5}u^5, \qquad b^2 v^3 \leq \frac{2}{5}b^5+\frac{3}{5}v^5$$
so by adding the two inequalities the claim $a^2 u^3 + b^2 v^3 \leq 1$ readily follows.