Math help Please

i think this is wrong....

or right...

using Lagrange multipliers you can show that the minimum occurs when a=b=c=d=1/4 and is equal to 4/3

but you can also argue this from symmetry since they are all required to be positive.

Let $x$, $y$, and $z$ be positive real numbers such that  $xyz=1$
Find the minimum value of $(x+2y)(y+2z)(xz+1)$.

$\mathbf{\huge{AM \geq GM }}$

$\begin{array}{|rcll|} \hline \dfrac{x+2y}{2} & \ge & \sqrt{x*2y} \\\\ \dfrac{y+2z}{2} & \ge & \sqrt{y*2z} \\\\ \dfrac{xz+1}{2} & \ge & \sqrt{xz*1} \\\\ \hline \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{x*2y}\sqrt{y*2z} \sqrt{xz*1} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{x*2y*y*2z*xz*1} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{4x^2y^2z^2} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & 2xyz \\\\ (x+2y)(y+2z)(xz+1) & \ge & 16*xyz \quad | \quad xyz=1 \\\\ (x+2y)(y+2z)(xz+1) & \ge & 16*1 \\\\ \mathbf{(x+2y)(y+2z)(xz+1)} & \ge & \mathbf{16} \\ \hline \end{array}$

The minimum value of $(x+2y)(y+2z)(xz+1)$ is $\mathbf{16}$