Hi, I'm a little stuck on this question.

Part a) asks to prove that $\displaystyle

\[l = \sqrt{a^{2}+b^{2}+c^{2}}\]$, where $\displaystyle \[l \]$ is the diagonal of a rectangular brick with sides of length a, b and c.

I've already proved that, by saying:

$\displaystyle \[d = \sqrt{a^{2}+c^{2}}\]$

$\displaystyle \[l = \sqrt{d^{2}+b^{2}}\]$

$\displaystyle \[l = \sqrt{\sqrt{a^{2}+c^{2}}^{2} + b^{2}}\]$

$\displaystyle \[\therefore l = \sqrt{a^{2}+b^{2}+c^{2}}\]$

The question I'm struggling with is part b), which says "If V denotes the volume of the brick, show that $\displaystyle \[V \leq (3\sqrt{3})^{-1}\ \times\ l^{3}\]

$.

I've found the AM by letting $\displaystyle a(1) =\ a^{2},\ a(2) =\ b^{2},\ a(3) =\ c^{2}$ then dividing by 3 and raising to the power of $\displaystyle \[\frac{3}{2}\]

$ to get $\displaystyle \[(3\sqrt{3})^{-1}\ \times\ l^{3}\]$.

I'm not sure what to do when finding the AM/GM inequality as I can't find a way to make GM = the volume (which is the only way I can think to solve this at the moment).

So far I've got $\displaystyle \[\frac{l^{3}}{3\sqrt{3}} = AM \geq GM = \sqrt[3]{a^{2}\times b^{2}\times c^{2}}\]

$, but I can't make any progress from there.

Is it as simple as $\displaystyle \[V =\ a\times b\times c \leq \sqrt[3]{a^{2}+b^{2}+c^{2}} \leq \frac{l^{3}}{3\sqrt{3}}\]$? Just seems to me like there is a neater solution and I've made some sort of formulaic or calculative error.

Appreciate any input.