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What is the minimum value of $p^2x + q^2y + r^2z$ if $pqxyz = 54r$, where $p, q, r, x, y$ and $z$ are positive real numbers?

I tried applying Cauchy's here but it didn't yield any significant result. Please help.

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Try to keep the title short and explain the question in the post. – user17762 Jun 30 '12 at 18:24
Are any of the variables fixed, or are they all allowed to vary? – mrf Jun 30 '12 at 19:50
up vote 2 down vote accepted

The problem might be incorrect since as such $p^2x + q^2 y + r^2z$ can be made arbitrarily close to $0$. Also, I assume that $pqxyz = 54r$ as your title says and not $pqxyz = 54c$ as your question says.

A way to see that is to take $p = \dfrac1x$, $q = \dfrac1y$ and $z = 54r$. The objective function now becomes $\dfrac1x + \dfrac1y + 54 r^3$. Now let $x,y \to \infty$ and $r \to 0$ to see that the objective function can be made arbitrary small.

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Sorry sir, its 54r only. – Bazinga Jun 30 '12 at 18:15

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