Trigonometric inequality

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Source: Baltic Way 2005/2

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Arne

#1 h Nov 7, 2005, 8:34 PM • 3 Y

Y by Adventure10, sohere, Mango247

Let $\alpha$ , $\beta$ and $\gamma$ be three acute angles such that $\sin \alpha+\sin \beta+\sin \gamma = 1$ . Show that
$\tan^{2}\alpha+\tan^{2}\beta+\tan^{2}\gamma \geq \frac{3}{8}.$

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nttu

#2 h Nov 8, 2005, 5:05 AM • 5 Y

Y by Centralorbit, Adventure10, sohere, Mango247, and 1 other user

We have : $tan^2\alpha +1 =\frac{1}{1-sin^2\alpha}$
The problem is equivalent to :
$\frac{1}{1-sin^2\alpha}+\frac{1}{1-sin^2\beta}+\frac{1}{1-sin^2\gamma} \geq \frac{27}{8}$
By Cauchy-Scharwz's ineq , we have :
$\frac{1}{1-sin^2\alpha}+\frac{1}{1-sin^2\beta}+\frac{1}{1-sin^2\gamma} \geq \frac{9}{3-(sin^2\alpha+sin^2\beta+sin^2\gamma)}$
It's enough to show that : $sin^2\alpha+sin^2\beta+sin^2\gamma \geq \frac{1}{3}$ , which is true by Cauchy again .

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#3 h Nov 8, 2005, 3:33 PM • 2 Y

Y by Adventure10, Mango247

by cauchy we have (cos²a+ cos²b+ cos²c) . (tg²a+ tg²b + tg²c) >= (sina +sinb +sinc)² = 1.
see that, again by cauchy, 3.(sin²a + sin²b + sin²c) >= (sina + sinb+ sinc)² = 1
this implies that 3- (cos²a + cos²b + cos²c)>= 1/3
so, cos²a+cos²b+cos²c <=3/8.

so, 3/8 . (tg²a+tg²b+tg²a) >= 1, and the result follow.

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#4 h Nov 8, 2005, 6:11 PM • 1 Y

Y by Adventure10

My solution is similar to nntu. put $sin a=x$ etc. So the problem is equivalent to.

$x,y,z>0$ $x+y+z=1$ and we want to prove that $\sum\frac{x^2}{1-x^2}\geq\frac{3}{8}$ .

We use cauchy in the engel form to have

$\sum\frac{x^2}{1-x^2}\geq\frac{(x+y+z)^2}{3-(x^2+y^2+z^2)}$

so it emains to prove that $\frac{(x+y+z)^2}{3-(x^2+y^2+z^2)}\geq\frac{3}{8}$

<=>...<=> $3(x^2+y^2+z^2)\geq 1$ which is true.

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