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.

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 $ . 
