## Monday, May 31, 2010

### Triangle Inequality

Given a triangle $ABC$ and a point $M$ inside the triangle.

Let $\alpha = \angle BMC, \beta = \angle AMC, \gamma = \angle AMB$

Prove that:
$$\frac{AM}{BM.CM} + \frac{BM}{CM.AM} + \frac{CM}{AM.BM} \geq -2 \left( \frac{\cos \alpha}{AM} + \frac{\cos \beta}{BM} + \frac{\cos \gamma}{CM} \right)$$

Solution In Progress

Let
$a = \frac{AM}{\sin \alpha}, b = \frac{BM}{\sin \beta}, c = \frac{CM}{\sin \gamma}$

Because $M$ is in the interior of the triangle, then $0 < \alpha, \beta, \gamma < \pi$ and thus $0 < \sin \alpha, \sin \beta, \sin \gamma \leq 1$. Thus $a,b,c > 0$. Without loss of generality, we may assume that $a \geq b \geq c$.

So we have:
$AM = a \sin \alpha, BM = b \sin \beta, CM = c \sin \gamma$

Substitute it to our inequality, and use the following shorthand:

$C_\alpha = \cos \alpha \sin \beta \sin \gamma$
$C_\beta = \sin \alpha \cos \beta \sin \gamma$
$C_\gamma = \sin \alpha \sin \beta \cos \gamma$

So our inequality becomes
$$\iff a^2\sin^2 \alpha + b^2 \sin^2 \beta + c^2 \sin^2 \gamma +2 ( bc C_\alpha + ac C_\beta + ab C_\gamma ) \geq 0$$

Note the following identities:
$$C_\beta + C_\gamma = \sin \alpha \cos \beta \sin \gamma + \sin \alpha \sin \beta \cos \gamma = \sin \alpha \sin (\beta + \gamma) = \sin \alpha \sin (2 \pi - (\beta + \gamma)) = - \sin^2 \alpha$$

Similarly,
$$C_\alpha + C_\gamma = - \sin^2 \beta$$
$$C_\alpha + C_\beta = - \sin^2 \gamma$$

So that
$$C_\alpha = (\sin^2 \alpha - \sin^2 \beta - \sin^2 \gamma)/2$$
$$C_\beta = (\sin^2 \beta - \sin^2 \alpha - \sin^2 \gamma)/2$$
$$C_\gamma = (\sin^2 \gamma - \sin^2 \beta - \sin^2 \alpha)/2$$

Substituting back to our inequalities, we have:
$$\iff (a-b)(a-c)\sin^2 \alpha + (b-a)(b-c) \sin^2 \beta + (c-a)(c-b) \sin^2 \gamma \geq 0$$

It's also equivalent to:
$$\iff (a-b)^2 C_\gamma + (a-c)^2 C_\beta + (b-c)^2 C_\alpha \leq 0$$