Minimize $(x+y)(y+z)(z+x)$ given $xyz(x+y+z) = 1$ $x,y,z$ are positive reals and I am given  $xyz(x+y+z) = 1$. Need to minimize $(x+y)(y+z)(z+x)$. Here is my approach.
Using AM-GM inequality
$$ (x+y) \geqslant 2 \sqrt{xy} $$
$$ (y+z) \geqslant 2 \sqrt{yz} $$
$$ (z+x) \geqslant 2 \sqrt{zx} $$
So, we have
$$ (x+y)(y+z)(z+x) \geqslant 8xyz $$
Also, I got
$$ \frac{x+y+z+(x+y+z)}{4} \geqslant \bigg[ xyz(x+y+z) \bigg] ^{1/4} $$
$$ \therefore x+y+z \geqslant 2 $$
But, I am stuck here. Any hints ?

$(x+y)(y+z)(z+x)=(z+x)(y(x+y+z)+xz)=(\frac{1}{zx}+zx)(x+z)$
now we can use $$\frac{1}{zx}+zx\ge 4{(\frac{1}{27{(xz)}^2})}^{1/4}$$
(HINT:$\frac{1}{zx}=\frac{1}{3zx}+\frac{1}{3zx}+\frac{1}{3zx}$)
also we can use $$x+z\ge 2\sqrt{xz}$$
Multiplying we get $$(x+y)(y+z)(z+x)\ge \frac{8}{3^{3/4}}$$
