Given that $x$, $y$ and $z$ are positive real numbers satisfying $xyz=32$, find the minimum value of:
Perhaps AM-GM and manipulation but I'm not quite sure how?
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Less elegantly, but more generally, you can also use a Lagrange multiplier, ie, minimize $$x^2+4xy+4y^2+2z^2 - \lambda(xyz-32)$$ as a function of $x,y,z$ and $\lambda$.
Solving the set of four coupled equations gives again $x=z=2y \;(=4)$, and the minimum value, 76, as well as the multiplier $\lambda = 2$.