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 ?
 A: $(x+y)(y+z)(z+x)=(z+x)(y(x+y+z)+xz)=(\frac{1}{zx}+zx)(x+z)$
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 $$(x+y)(y+z)(z+x)\ge \frac{8}{3^{3/4}}$$
A: 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 $xyz(x+y+z)=1$
A: 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 $x=y=z=\frac{1}{\sqrt[4]{3}}.$
A: 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!
