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 :) 
 A: 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 $S\le 4$.
A: Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$.
Hence, since $(x,y,z,t)$ and $(xyz,xyt,xzt,yzt)$ are the same ordered,
by Rearrangement and AM-GM we obtain:
$$a^2bc+b^2cd+c^2da+d^2ab=a\cdot abc+b\cdot bcd+c\cdot cda+d\cdot dab\leq$$
$$\leq x\cdot xyz+y\cdot xyt+z\cdot xzt+t\cdot yzt=xy(xz+yt)+zt(xz+yt)=$$
$$=(xy+zt)(xz+yt)\leq\left(\frac{xy+xz+zt+yt}{2}\right)^2=$$
$$=\left(\frac{(x+t)(y+z)}{2}\right)^2\leq\left(\frac{\left(\frac{x+y+z+t}{2}\right)^2}{2}\right)^2=4.$$
Done!
