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

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

Perhaps AM-GM and manipulation but I'm not quite sure how?

Source BMO.

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    $\begingroup$ isn't this a bit too easy for an IMO problem? Which IMO is it? $\endgroup$ – Albanian_EAGLE Apr 14 '13 at 2:11
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    $\begingroup$ Sorry, Typo, not IMO but BMO, British Maths Olympiad and an early one, these days it is much harder. $\endgroup$ – Dominic Stone Apr 30 '13 at 8:56
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Yes AM-GM is the right approach.

By AM-GM,

$$\frac{x^2+2xy+2xy+4y^2+z^2+z^2}{6} \geq (16 \cdot x^4 \cdot y^4 \cdot z^4)^{\frac{1}{6}}=$$

$$=(2^{24})^{\frac{1}{6}}=16$$ with equality iff $x=z=2y$.

So the minimum value of $x^2+4xy+4y^2+2z^2$ is $16 \cdot 6=96$. QED

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  • $\begingroup$ Thanks so much, sometimes I find it hard to find the right way to rearrange such expressions. Beautiful solution. $\endgroup$ – Dominic Stone Apr 12 '13 at 5:34
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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 $\lambda = 2$.

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