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


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


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

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

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

    Now, $\displaystyle S>1$ $\displaystyle \iff$ $\displaystyle S^2>1$ $\displaystyle \iff$ $\displaystyle 8x^6+4x^4-12x^3-1>0.$ Thus if you can show that $\displaystyle x\geqslant3\ \Rightarrow\ 8x^6+4x^4-12x^3-1>0,$ 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,


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

    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; Oct 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 $\displaystyle S^2=\frac{(abc)^2(ab-2)(bc-2)(ca-2)}{16}$ (hopefully thatís correct this time). Then $\displaystyle S^2>1\ \Leftrightarrow\ (ab-2)(bc-2)(ca-2)>\frac{16}{(abc)^2}.$ However it is possible that this inequality may not be true for $\displaystyle (a,b,c)$ arbitrarily close to $\displaystyle \left(\sqrt2,\sqrt2,2\sqrt2\right).$ See this.
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  8. #8
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    This is about an acute triangle so $\displaystyle \alpha,\beta,\gamma<\frac {\pi}2$...
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