Simple symmetric inequality involving four variables

Question

This problem was a part of the AM-GM section, but you are free to use any inequality that crosses your mind. Given is that $a,b,c,d >0$ and $a+b+c+d=4$. To prove:

$$ \frac{4}{abcd} \geq \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} $$

Answer 1

We need to prove that $a^2cd+b^2ad+c^2ac+d^2bc\leq4$.

Let $\{a,b,c,d\}=\{x,y,z,t\}$ such that $x\geq y\geq z\geq t$.

Hence, by Rearrangement and AM-GM $$a^2cd+b^2ad+c^2ab+d^2bc=$$ $$=a\cdot acd+b\cdot abd+c\cdot abc+d\cdot bcd\leq$$ $$\leq x\cdot xyz+y\cdot xyt+z\cdot xzt+t\cdot yzt=$$ $$=(xy+zt)(xz+yt)\leq\left(\frac{xy+zt+xz+yt}{2}\right)^2=$$ $$=\left(\frac{(x+t)(y+z)}{2}\right)^2\leq\left(\frac{\left(\frac{x+y+z+t}{2}\right)^2}{2}\right)^2=4$$ Done!