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

Question

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 :)

Answer 1

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 $S\le 4$.