# Prove the inequality $a^2bc+b^2cd+c^2da+d^2ab \leq 4$ with $a+b+c+d=4$

Let $a,b,c$ and $d$ be positive real numbers such that $a+b+c+d=4.$

Prove the inequality

$$a^2bc+b^2cd+c^2da+d^2ab \leq 4 .$$

Thanks :)

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@Iuli: as you already know, the inequalities may be a painful thing since some of them require a lot of time to think. That's why it helps mention anything you know about it, what you've already tried. You have (+1) for the question. Could Cauchy-Schwarz help? – user 1618033 Aug 19 '12 at 13:24
It may be worth noting that the LHS can be written as $abcd\left(\frac{b}{a}+\frac{a}{b}+\frac{d}{c}+\frac{c}{d}\right)$ – Daniel Littlewood Aug 19 '12 at 15:49

## 1 Answer

Let $S=a^2bc+b^2cd+c^2da+d^2ab$. We can easily find that:

$$S-(ac+bd)(ab+cd)=-bd(a-c)(b-d);$$ $$S-(bc+ad)(bd+ac)=ac(a-c)(b-d)$$ which implies $$S\le \max\{(ac+bd)(ab+cd),(bc+ad)(bd+ac)\}.$$

By AG mean inequality:

\begin{align*} (ac+bd)(ab+cd)&\le \left(\frac{(ac+bd)+(ab+cd)}{2}\right)^2\\ {}&=\frac{(a+d)^2(b+c)^2}{4}\\ {}&\le \frac{1}{4}\left[\left(\frac{a+d+b+c}{2}\right)^2\right]^2\\ {}&=4 \end{align*} Similarly, we have $$(bc+ad)(bd+ac)\le 4$$.

Thus we have $S\le 4$.

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+1 Quite neat!${}{}{}{}$ – Sasha Aug 19 '12 at 16:30