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If $x,y,z$ are positive real no. and $xyz= 32\;,$ Then Minimum value of $$x^2+4xy+4y^2+4z^2$$ is

$\bf{My\; Try::}$ Here I have Used $\bf{A.M\geq G.M}$ Inequality

So $$\displaystyle \frac{x^2+4xy+4y^2+4z^2}{4}\geq \left(x^3\cdot y^3\cdot z^2\right)^{\frac{1}{4}}\;,$$ But I did not How can I solve it

Help me, Thanks

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Applying the AM-GM is the right strategy, but you need to do it a bit differently.

$$x^2+4xy+4y^2+4z^2$$

$$= x^2+2xy+2xy+4y^2+2z^2+2z^2$$

$$\ge 6\sqrt[6]{x^2 \cdot 2xy \cdot 2xy \cdot 4y^2 \cdot 2z^2 \cdot 2z^2}$$

$$= 6\sqrt[6]{64x^4y^4z^4}$$

$$= 12(xyz)^{2/3}$$

$$= 12 \cdot 32^{2/3}$$

Equality holds when $x^2 = 2xy = 4y^2 = 2z^2$ and $xyz = 32$, which is attained when $(x,y,z) = (2^{13/6},2^{7/6},2^{10/6})$.

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Lagrange's theorem is helpful.

Let $f: (x,y,z) \mapsto x^{2} + 4xy + 4y^{2} + 4z^{2}$ on $\mathbb{R}^{3}$; let $g: (x,y,z) \mapsto xyz - 32$ on $\mathbb{R}^{3}_{++}$; and let $(x,y,z) \in g^{(-1)}\{ 0 \}$ such that $f(x,y,z)$ is an extremum. Then there is some $\lambda \in \mathbb{R}$ such that $$ \nabla f(x,y,z) = (2x+4y, 4x+8y, 8z) = \lambda \nabla g = \lambda (yz, xz, xy). $$ And we can find the candidate possible extrema of $f|_{g^{(-1)}\{ 0 \}}$, inspecting them.

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    $\begingroup$ The tag is (algebra-precalculus), though. Perhaps something different is expected. $\endgroup$ – Brian Tung Aug 13 '15 at 6:55

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