1. ## Challenging trigonometry problem

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

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

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

4. Originally Posted by mestmoha
I tried many expansions! I just could not prove it...

5. ## Challenging trigonometry problem

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,

6. 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?

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

8. 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 : $\frac{1-\sqrt{5}}{2} \quad and \quad \frac{1+\sqrt{5}}
{2}$

Hence:

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

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 :

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