Prove inequality $ \frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12 $ Let $x,y,z > 0$ and $xyz=8.$ Prove that 
$$
\frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12
$$
 I have tried with AM–GM inequality but no result.

We'll prove that $\sum\limits_{cyc}\frac{(x+y-1)^2}{z}\geq\frac{27}{2}$.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $w=2$ and we need to prove that $f(v^2)\geq0$, where
$f(v^2)=9u^2v^2-5uw^3-3uv^2w^3-\frac{5}{4}w^4+\frac{1}{4}v^2w^2$.
But $f$ is a linear function, which says that it's enough to prove our inequality 
for an extremal value of $v^2$, which happens for equality case of two variables.
Indeed, $x$, $y$ and $z$ are positive roots of the equation $(X-x)(X-y)(X-z)=0$ or
$X^3-3uX^2+3v^2X-w^3=0$ or $3v^2X=-X^3+3uX^2+w^3$.
Thus, the last equation has three positive roots, which says that a line $Y=3v^2X$ 
and a graph of $Y=-X^3+3uX^2+w^3$ have three common points (draw it!),
which says that $v^2$ gets an extremal value, when a graph of $Y=3v^2X$ 
is a tangent line to the graph of $Y=-X^3+3uX^2+w^3$,
which happens for equality case of two variables. 
Let $y=x$ and $z=\frac{8}{x^2}$. 
We obtain $(x-2)^2(4x^7+12x^6+33x^5+100x^4+128x^3+128x^2+256x+256)\geq0$.
Done!
