<s>[INST] 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 ?
 [/INST] Setting $t=xyz(x+y+z),$ then $xy+yz+zx \geqslant \sqrt{3t}=\sqrt 3.$ Using known inequality
$$(x+y)(y+z)(z+x) \geqslant \frac{8}{9}(x+y+z)(xy+yz+zx).$$
We have
$$(x+y)(y+z)(z+x) \geqslant \frac{8}{9}(x+y+z)(xy+yz+zx)$$
$$\geqslant \frac{8}{9} \cdot \sqrt{3(xy+yz+zx)} \cdot (xy+yz+zx) = \frac{8}{9}\sqrt{3(xy+yz+zx)^3}$$
$$\geqslant \frac{8}{9}\sqrt{3\left(\sqrt{3}\right)^3} = \frac{8\sqrt[4]{3}}{3}.$$
Equality occur when $x=y=z=\frac{1}{\sqrt[4]{3}}.$
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