Challenging trigonometry problem

**mestmoha** · February 22nd 2011, 09:23 AM

Please, I need help to prove:

$(\sin A + \sin B \sin C)^2 \leq 1 + \sin A + \sin B \sin C$
such that A, B, and C are triangle angles

Thank you

**pickslides** · February 22nd 2011, 12:06 PM

Have you expanded out the LHS? Can you apply any identities here?

**mestmoha** · February 22nd 2011, 12:23 PM

I tried many expansions! I just could not prove it...

**mr fantastic** · February 22nd 2011, 05:20 PM

Originally Posted by mestmoha

I tried many expansions! I just could not prove it...

Please post all your working.

**mestmoha** · February 22nd 2011, 06:01 PM

we should prove:
$\alpha = 1 + \sin A + \sin B \sin C - (\sin A + \sin B \sin C)^2 \geq 0$
we know that:

$\frac{\sin A}{a} = \frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}$
such that R is the cimscribed circle radius

by Substituting in alpha, we need to prove now:

$\frac{16 R^4 + 8 aR^3 + 4 bc R^2 - (4 a^2 R^2 + 4 abc R + b^2 c^2) } {16 R^4} \geq 0$

Such that a, b,and c are the edges of a triangle.

As a remark,

$R=\frac{abc}{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}}$

Thank you,

**scounged** · February 23rd 2011, 01:57 PM

Originally Posted by mestmoha

we should prove:
$\alpha = 1 + \sin A + \sin B \sin C - (\sin A + \sin B \sin C)^2 \geq 0$

Where did alpha come from? Did you forget to write it out in your first post?

**mestmoha** · February 23rd 2011, 02:24 PM

Alpha does not change anything, it is just the difference between the LHS and the RHS

**mestmoha** · February 27th 2011, 12:07 PM

Originally Posted by mestmoha

Alpha does not change anything, it is just the difference between the LHS and the RHS

I think I found the solution. Please, check it for me:
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
$\frac{{\sin}^2 A}{{a}^2}=\frac{\sin B \sin C}{bc}$
$\sin A + \sin B \sin C= \Sin A (1 + \frac{bc}{{a}^2} \sin A)$

For any A, and a $\sin A + \sin B \sin C$ is maximum if and only if b=c (you can use the area= 1/2 bc sin A to approach it).

hence for a triangle with b=c:

$\sin A + \sin B \sin C= \sin A + \frac{1}{2} \cos A +\frac{1}{2}$

if:
$x=\sin A \quad then \quad f(x) =x + \frac{1}{2}\sqrt{1-x^2} + \frac {1} {2}$
by using the derrivative of f(x), and setting f'(x)=0 we get:
$x=\sin A= \frac{2}{\sqrt{5}}$

Then the maximum value of f(x) $\sin A + \sin B \sin C \quad$ ) is $\frac{1+\sqrt{5}}{2}$

now we proved that for a triangle ABC $\sin A + \sin B \ sin C \leq \frac{1 + \sqrt{5}}{2}$

Let's consider a function g(x) such that:
$g(x)=-x^2 + x +1$

The roots of g(x) are : $\frac{1-\sqrt{5}}{2} \quad and \quad \frac{1+\sqrt{5}}<br /> {2}$

Hence:

$for \quad every \quad x \quad: \frac{1-\sqrt{5}}{2} \leq x \leq \frac{1+\sqrt{5}}<br /> {2}$

and since :

$0\leq \sin A + \sin B \sin C \leq \frac {1+\sqrt{5}}{2}$ (as we proved)

Then:

$g(\sin A + \sin B \sin C) \geq 0$

-->

$- {(\sin A + \sin B \sin C)}^{2} + (\sin A + \sin B \sin C) +1 \geq 0$

Thus :

$(\sin A + \sin B \sin C)}^{2} \leq (\sin A + \sin B \sin C) +1$

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