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 :) 

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!
