<s>[INST] 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 :) 
 [/INST] 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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