# Thread: Graph Theory: Arithmetic and Geometric Means

1. ## Graph Theory: Arithmetic and Geometric Means

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.

2. Your first step was correct:
Let $\displaystyle a(1) =\ a^{2},\ a(2) =\ b^{2},\ a(3) =\ c^{2}$

AM = $\displaystyle \frac {a^{2} + b^{2} + c^{2}}{3}$

GM = $\displaystyle (a^{2}b^{2}c^{2})^{\frac{1}{3}}$

By AM GM inequality,

$\displaystyle (a^{2}b^{2}c^{2})^{\frac{1}{3}} \leq \frac {a^{2} + b^{2} + c^{2}}{3}$

Does taking square roots on both sides bring the LHS and the RHS into some known form ?

3. Ahh great, got it with a bit of messing around after that push in the right direction.

For those interested: Taking the square roots gives $\displaystyle $\sqrt[3]{abc}\leq \frac{\sqrt{3}\times \sqrt{a^{2}+b^{2}+c^{2}}}{3} = \frac{l\sqrt{3}}{3}$$

From there,

$\displaystyle $3\sqrt[3]{abc}\leq l\sqrt{3}$$
$\displaystyle $\sqrt{3}\sqrt[3]{abc}\leq l$$
$\displaystyle $3\sqrt{3} abc\leq l^{3}$$
$\displaystyle $abc = V\leq \frac{l^{3}}{3\sqrt{3}}$$

Thanks Traveller!