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

**james_bond** · October 27th 2009, 12:21 AM

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 $S=\frac {abc}{4R}=\frac {\sin\alpha+\sin\beta+\sin\gamma}2\le \frac {3\sqrt {3}/2}2$ ).

**proscientia** · October 27th 2009, 01:58 AM

What is $S\,?$

**james_bond** · October 27th 2009, 03:35 AM

The area of the triangle.

**proscientia** · October 27th 2009, 03:55 AM

By Heron’s formula for the area of the triangle,

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

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

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

**james_bond** · October 27th 2009, 12:53 PM

Originally Posted by proscientia

By Heron’s formula for the area of the triangle,

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

Sorry for not noticing sooner but your solution appears to be incorrect. By Heron's formula:
$S=\sqrt{\frac {abc}2\left(\frac {abc}2-a\right)\left(\frac {abc}2-b\right)\left(\frac {abc}2-c\right)}=\frac {abc}4\sqrt{(-2+a b) (-2+a c) (-2+b c)}$ $=\frac {abc}4\sqrt{(abc)^2-2abc(a+b+c)+4(ab+ac+bc)-8}=$ $\frac {abc}4\sqrt{-(abc)^2+4(ab+ac+bc)-8}$ $\therefore S^2\geqslant \frac {(abc)^2}{16}\left(12(abc)^{\frac 23}-(abc)^2-8\right)$ 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.

**proscientia** · October 28th 2009, 09:14 AM

Yes, I’ve just noticed that. It was very careless of me. Sorry.

**proscientia** · October 29th 2009, 03:24 AM

Hmm, I wondering if what is to be proved is actually false. Heron’s formula gives $S^2=\frac{(abc)^2(ab-2)(bc-2)(ca-2)}{16}$ (hopefully that’s 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.

**james_bond** · October 29th 2009, 08:48 AM

This is about an acute triangle so $\alpha,\beta,\gamma<\frac {\pi}2$ ...

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