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 ?

For $x=y=z=\frac{1}{\sqrt[4]3}$ we get a value $\frac{8}{\sqrt[4]{27}}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that $$\prod_{cyc}(x+y)\geq\frac{8}{\sqrt[4]{27}}$$ or
$$27\prod_{cyc}(x+y)^4\geq4096x^3y^3z^3(x+y+z)^3.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that $$(9uv^2-w^3)^4\geq4096u^3w^9$$ or $f(w^3)\geq0,$ where
$$f(w^3)=(9uv^2-w^3)^4-4096u^3w^9.$$
But it's obvious that $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $w^3$, which by $uvw$ happens for equality case of two variables.
Since the last inequality is symmetric and homogeneous, it's enough to assume $y=z=1$ and we need to prove that:
$$27(x+1)^8\geq256x^3(x+2)^3,$$ which is true by AM-GM:
$$27(x+1)^8=27(x^2+2x+1)^4=27\left(3\cdot\frac{x^2+2x}{3}+1\right)^4\geq$$
$$\geq27\left(4\sqrt[4]{\left(\frac{x^2+2x}{3}\right)^3\cdot1}\right)^4=256x^3(x+2)^3$$ and we are done!
