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.
 A: Using the Cauchy-Scwarz inequality 
$$ (a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}) \geq (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^{2} $$ with $a_{1}=\frac{x+y-1}{\sqrt{z}} $, $ a_{2}=\frac{x+z-1}{\sqrt{y}} $ $ a_{3}=\frac{z+y-1}{\sqrt{x}} $ and $ b_{1}=\sqrt{z} $, $ b_{2}=\sqrt{y} $ and $ b_{3}=\sqrt{x} $, we obtain that  $$\Big (\frac{(x+y-1)^{2}}{z}+\frac{(x+z-1)^{2}}{y}+\frac{(z+y-1)^{2}}{x} \Big) \cdot S \geq (2S-3)^{2} $$
where $ S=x+y+z $. 
Therefore it suffices to show that $\frac{(2S-3)^{2}}{S} \geq 12 $ which is equivalent to showing that $ S \in (-\infty, \frac{6-3\sqrt{3}}{2}]\cup [ \frac{6+3\sqrt{3}}{2},\infty ) $. 
But we have that $ S=x+y+\frac{8}{xy} $ and using the AM-GM, this is greater than  or equal to $ 3(xy\cdot 8\frac{1}{xy})^{\frac{1}{3}}=6 $. 
since clearly $ 6> \frac{6+3\sqrt{3}}{2} $, the conclusion follows.
Moreover, $ \frac{27}{2} $ is the sharpest lower bound and in the same manner we can prove that the LHS is $ \geq \frac{27}{2} $ since this is equivalent to $ S \in (-\infty, \frac{3}{8}] \cup [6, \infty) $ after computations and this is  true since we proved earlier that $ S \geq 6 $ . 
A: 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!
A: The solutions provided so far all start with some clever substitutions but, at least for me, I don't see how I find these in the first place. 
Hence let us look at a heuristic approach.
a) Because of the symmetry of the equations we assume the solution to be symmetric as well and we simply set all three variables equal to x. We leave a justifying of this assumption open here. Hence this is, strictly speaking, not a solution but an extended comment.
b) to find the minimum we use a little calculus and the method of the Langrange multiplicator
The function becomes
$$f_s =\frac{3 (2 x-1)^2}{x}$$
The supplementary condition becomes (we allow for a parameter q instead of 2 is in the OP)
$$g=0$$
where
$$g=x^3-q^3$$
Therefore the function to be minimized is
$$f_m = f_s+g \;\lambda$$
Where $\lambda$ is the Langrange multiplicator.     
Setting the derivatives with respect to $x$ and $\lambda$ equal to 0 and solving we obtain
$$\left\{x=q,\lambda =\frac{1-4 q^2}{q^4}\right\}$$
The minimum becomes
$$fmin = (3 (-1 + 2 q)^2)/q$$
