Given that $x, y, z$ are positive real numbers satisfying $xyz = 32$, find the minimum value of

\begin{align*} x^{2} + 4xy + 4y^{2} + 2z^{2} \end{align*}

### Solution

Using $AM \geq GM$ as all the quantities involved are positive, we have

\begin{align*} \frac{x^{2} + 2xy + 2xy + 4y^{2} + z^{2} + z^{2}}{6} &\geq& (x^{2} \cdot 2xy \cdot 2xy \cdot 4y^{2} \cdot z^{2} \cdot z^{2})^{\frac{1}{6}} = 16 \ \implies x^{2} + 4xy + 4y^{2} + 2z^{2} &\geq& 96 \end{align*}

Therefore, the minimum value of $x^{2} + 4xy + 4y^{2} + 2z^{2}$ is $96$.