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Math Help - Graph Theory: Arithmetic and Geometric Means

  1. #1
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    Graph Theory: Arithmetic and Geometric Means

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

    Part a) asks to prove that  <br />
\[l = \sqrt{a^{2}+b^{2}+c^{2}}\], where \[l \] is the diagonal of a rectangular brick with sides of length a, b and c.

    I've already proved that, by saying:
    \[d = \sqrt{a^{2}+c^{2}}\]
    \[l = \sqrt{d^{2}+b^{2}}\]
    \[l = \sqrt{\sqrt{a^{2}+c^{2}}^{2} + b^{2}}\]
    \[\therefore l = \sqrt{a^{2}+b^{2}+c^{2}}\]

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

    I've found the AM by letting a(1) =\ a^{2},\ a(2) =\ b^{2},\ a(3) =\ c^{2} then dividing by 3 and raising to the power of \[\frac{3}{2}\]<br />
to get \[(3\sqrt{3})^{-1}\ \times\ l^{3}\].

    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 \[\frac{l^{3}}{3\sqrt{3}} = AM \geq  GM = \sqrt[3]{a^{2}\times b^{2}\times c^{2}}\]<br />
, but I can't make any progress from there.

    Is it as simple as \[V =\ a\times b\times c \leq \sqrt[3]{a^{2}+b^{2}+c^{2}} \leq \frac{l^{3}}{3\sqrt{3}}\]? 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; September 22nd 2010 at 05:00 AM.
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  2. #2
    Member Traveller's Avatar
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    Your first step was correct:
    Let a(1) =\ a^{2},\ a(2) =\ b^{2},\ a(3) =\ c^{2}

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

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

    By AM GM inequality,

    (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 ?
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  3. #3
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    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 \[\sqrt[3]{abc}\leq \frac{\sqrt{3}\times \sqrt{a^{2}+b^{2}+c^{2}}}{3} = \frac{l\sqrt{3}}{3}\]

    From there,

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

    Thanks Traveller!
    Last edited by domislong; September 22nd 2010 at 05:35 AM.
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