TITLE: Given that $x,y,z$ are positive reals such that $xyz=32$.What is the minimum value of $x^2+4xy+4y^2+2z^2$.
QUESTION [5 upvotes]: Given that $x,y,z$ are positive reals such that $xyz=32$.What is the minimum value of $x^2+4xy+4y^2+2z^2$.

I could solve this question by Lagrange Multiplier method.And i found the minimum value of the expression as $96$.
I tried other methods to find the minimum value like AM-GM inequality,but i could not solve it.Can someone please tell me a method other than Lagrange Multiplier to find its minimum value.

REPLY [9 votes]: It is a matter of arranging terms to get $xyz$ right.  Using AM-GM,
$$x^2+2xy+2xy+4y^2+z^2+z^2\ge 6\sqrt[6]{16x^4y^4z^4}=96$$
As equality is possible when $x=2y=z=4$ we have the minimum.