Results 1 to 3 of 3

Thread: Graph Theory: Arithmetic and Geometric Means

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    2

    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.
    Last edited by domislong; Sep 22nd 2010 at 05:00 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Traveller's Avatar
    Joined
    Sep 2010
    Posts
    162
    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 ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2010
    Posts
    2
    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!
    Last edited by domislong; Sep 22nd 2010 at 05:35 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: Dec 14th 2011, 05:02 AM
  2. Replies: 3
    Last Post: Apr 30th 2011, 01:14 AM
  3. Replies: 4
    Last Post: Nov 9th 2010, 01:45 PM
  4. inequality relating to arithmetic and geometric means
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Dec 9th 2009, 08:51 PM
  5. Arithmetic/Geometric Means
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Sep 17th 2006, 11:43 AM

Search Tags


/mathhelpforum @mathhelpforum