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Let $$f(r)=\frac{1}{r}(\frac{4V}{\pi}+2V) + \pi r^2$$ $V$ stands volume and $r$ stands for radius

I need to find the extrema of the expression and find the $r$ at which the extrema occurs. So I used $$\frac{\frac{1}{r}(\frac{4V}{\pi})+\frac{1}{r}(2V) + \pi r^2}{3} \geq ({(4V)(2V) \pi})^{\frac{1}{3}}$$

However, my textbook applied the AM-GM inequality as

$$\frac{\frac{1}{r}(\frac{2V}{\pi}+V)+ \frac{1}{r}(\frac{2V}{\pi}+V) + \pi r^2}{3} \geq ((\frac{2V}{\pi}+V)^2\pi)^{\frac{1}{3}}$$.

Which method is the correct one? And how to find the the $r$ at which $f(r)$ is extremum ?

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    Remember that the AM-GM inequality is an equality iff all terms are the same. Now, as you have applied the inequality, it's impossible for $4V/\pi r$ to equal $2V/r$, so while the lower bound is valid, you'll also never reach it. The book circumvents this by putting it in the given form. – 2017-02-11
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    @stochasticboy321 You should consider posting that as an answer. – 2017-02-11

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