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 ?

Since $x+y+z \geqslant 3 \sqrt[3]{xyz}$, we have $xyz (x+y+z)\geqslant 3 (xyz)^{4/3}$.
Using the given condition, we have $1 \geqslant 3 (xyz)^{4/3}$. This is $xyz \leqslant \frac{1}{3^{3/4}} $
Also, we have $(x+y)(y+z)(z+x) = (x+y+z)(xy+ yz + zx) - xyz $
Now $ -xyz \geqslant - \frac{1}{3^{3/4}} $ and
$$ (xy+ yz + zx) \geqslant  3 (xyz)^{2/3} $$
$$ (x+y+z)(xy+ yz + zx) \geqslant 3 \frac{(x+y+z)xyz}{(xyz)^{1/3}} $$
$$ (x+y+z)(xy+ yz + zx) \geqslant  \frac{3}{(xyz)^{1/3}} $$
But, we have
$$ xyz \leqslant \frac{1}{3^{3/4}} $$
$$ \therefore \frac{1}{xyz} \geqslant 3^{3/4} $$
$$ \therefore \frac{1}{(xyz)^{1/3}} \geqslant 3^{1/4} $$
So, we get
$$ (x+y+z)(xy+ yz + zx) \geqslant  3^{5/4} $$
So, we have
$$(x+y+z)(xy+ yz + zx) -xyz \geqslant  3^{5/4} -  \frac{1}{3^{3/4}} $$
It follows that
$$ (x+y)(y+z)(z+x) \geqslant \frac{8}{3^{3/4}} $$
It can be seen that equality is achieved when $x=y=z$ in $xyz(x+y+z)=1$
