# Prove the inequality using AM-GM inequality

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.

• Just because you mentioned Holder, giving that way too: $$1\ge (a^5+b^5)^2(u^5+v^5)^3\ge(a^2u^3+b^2v^3)^5$$ – Macavity Jun 22 '15 at 16:52
• Another way would be to use Muirheads inequality (hated by olympiad graders) and once you have figured that out use weighted AM GM to get the same solution which is much preferred. This is what @user26486 did essentially – Faraz Masroor Jun 22 '15 at 22:01

## 2 Answers

Hint: by AM-GM:

$$5a^2u^3\le a^5+a^5+u^5+u^5+u^5$$

$$5b^2v^3\le b^5+b^5+v^5+v^5+v^5$$

Add these up:

$$5a^2u^3+5b^2v^3\le 2(a^5+b^5)+3(u^5+v^5)$$

Now use the conditions given.

• Out of curiosity, how long did it take to solve that? The telltale a^2 u^3 along with the AM-GM hint suggests we use AM-GM on {a^5,a^5,u^5,u^5,u^5} – smci Jun 22 '15 at 21:05
• @smci We prove $A\le C$ by proving $A\le B$ with $B\le C$ being a known/simple inequality. In this case, $$A=a^2u^3+b^2v^3\le \frac{a^5+a^5+u^5+u^5+u^5}{5}+\frac{b^5+b^5+v^5+v^5+v^5}{5}=B$$ with $B\le C=1$ given in the conditions. It's a very natural approach. – user26486 Jun 22 '15 at 21:11
• Yeah I understood, I said "use AM-GM on {a^5,a^5,u^5,u^5,u^5}". I merely asked how long it took to spot. – smci Jun 22 '15 at 21:13
• Also see Muirhead's inequality. – Faraz Masroor Jun 22 '15 at 22:01

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.

• Thanks for downvoting without even leaving a comment. I know the OP is supposed to use the AM-GM inequality, I just wanted to show that the Young's inequality is as much as effective. – Jack D'Aurizio Jun 22 '15 at 18:54
• The best answer ... +1 – user399078 Jan 31 '17 at 12:16