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Math Help - Prove that if abc=a+b+c in an acute triangle then 1<S

  1. #1
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    Prove that if abc=a+b+c in an acute triangle then 1<S

    Show that if abc=a+b+c in an acute triangle then the area of the triangle is greater than 1 ( 1<S). (I'm not sure if this involves trigonometry at all.)
    (I know that S\le \frac {3\sqrt{3}}4 because S=\frac {abc}{4R}=\frac {\sin\alpha+\sin\beta+\sin\gamma}2\le \frac {3\sqrt {3}/2}2).
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  2. #2
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    What is S\,?
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  3. #3
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    The area of the triangle.
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  4. #4
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    By Heron’s formula for the area of the triangle,


    \begin{array}{rcl}S &=& \sqrt{2abc(2abc-a)(2abc-b)(2abc-c)}\\\\<br />
{} &=& abc\sqrt{2(2ab-1)(2bc-1)(2ca-1)}\\\\<br />
{} &=& abc\sqrt{2\left[8(abc)^2-4abc(a+b+c)+2(ab+bc+ca)-1\right]}\\\\<br />
{} &=& abc\sqrt{2\left[4(abc)^2+2(ab+bc+ca)-1\right]}\end{array}


    By AM–GM, ab+bc+ca\geq3(abc)^{\frac23}.

    \therefore\ S^2\geqslant2(abc)^2\left[4(abc)^2+6(abc)^{\frac23}-1\right]=2x^3(4x^3+6x-1) where x=(abc)^{\frac23}.

    But, again by AM–GM, abc=a+b+c\geqslant3(abc)^{\frac13} \implies (abc)^2\geqslant27 \implies x\geqslant3.

    Now, S>1 \iff S^2>1 \iff 8x^6+4x^4-12x^3-1>0. Thus if you can show that x\geqslant3\ \Rightarrow\ 8x^6+4x^4-12x^3-1>0, you are done. I think this should be straightforward.
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  5. #5
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    Quote Originally Posted by proscientia View Post
    By Heron’s formula for the area of the triangle,


    \begin{array}{rcl}S &=& \sqrt{2abc(2abc-a)(2abc-b)(2abc-c)}\\\\<br />
{} &=& abc\sqrt{2(2ab-1)(2bc-1)(2ca-1)}\\\\<br />
{} &=& abc\sqrt{2\left[8(abc)^2-4abc(a+b+c)+2(ab+bc+ca)-1\right]}\\\\<br />
{} &=& abc\sqrt{2\left[4(abc)^2+2(ab+bc+ca)-1\right]}\end{array}
    Sorry for not noticing sooner but your solution appears to be incorrect. By Heron's formula:
    S=\sqrt{\frac {abc}2\left(\frac {abc}2-a\right)\left(\frac {abc}2-b\right)\left(\frac {abc}2-c\right)}=\frac {abc}4\sqrt{(-2+a b) (-2+a c) (-2+b c)} =\frac {abc}4\sqrt{(abc)^2-2abc(a+b+c)+4(ab+ac+bc)-8}= \frac {abc}4\sqrt{-(abc)^2+4(ab+ac+bc)-8} \therefore S^2\geqslant \frac {(abc)^2}{16}\left(12(abc)^{\frac 23}-(abc)^2-8\right) which implies -\frac{x^3}{2}+\frac{3 x^4}{4}-\frac{x^6}{16}>1 which is obviously false (for a general x(\geqslant 3)).

    I'm eagerly waiting for a good solution. (The upper was so simple to get I assume the lower should come just as easy but I can't see ) Any help would be greatly appreciated! Thanks.
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  6. #6
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    Yes, I’ve just noticed that. It was very careless of me. Sorry.
    Last edited by proscientia; October 28th 2009 at 02:09 PM.
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  7. #7
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    Hmm, I wondering if what is to be proved is actually false. Heronís formula gives S^2=\frac{(abc)^2(ab-2)(bc-2)(ca-2)}{16} (hopefully thatís correct this time). Then S^2>1\ \Leftrightarrow\ (ab-2)(bc-2)(ca-2)>\frac{16}{(abc)^2}. However it is possible that this inequality may not be true for (a,b,c) arbitrarily close to \left(\sqrt2,\sqrt2,2\sqrt2\right). See this.
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  8. #8
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    This is about an acute triangle so \alpha,\beta,\gamma<\frac {\pi}2...
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