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

Given that $x,y,z$ are positive reals such that $xyz=32$.What is the minimum value of $x^2+4xy+4y^2+2z^2$.

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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.

Answer 1

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.

Answer 2

You could substitute $x = \frac{32}{yz}$ and then get $f(y,z) = (\frac{32}{yz})^2+4\cdot 32/z + 4y^2+2z^2$. Then search the minimum of this function. link