Inequality in triangle

Question

Let $ABC$ be a triangle and $M$ a point on side $BC$. Denote $\alpha=\angle BAM$, $\beta=\angle CAM$. Is the following inequality true? $$\sin \alpha \cdot (AM-AC)+\sin \beta \cdot (AM-AB) \leq 0.$$

Answer 1

As I already stated in a comment, your question only makes sense if you assume $0\le\alpha,\beta\le\pi$. As a consequence you get $\sin\alpha\ge0,\sin\beta\ge0$ which will be very useful.

I use $a$ to denote the length $AB$ and $b$ to denote $AC$. Yes, the letters are a bit counter-intuitive with respect to the points $B$ and $C$, but they fit in well with $\alpha$ and $\beta$. Using $a,b,\alpha$ and $\beta$ as parameters, you can deduce

$$AM=\frac{ab\sin\alpha\cos\beta + ab\cos\alpha\sin\beta} {a\sin\alpha + b\sin\beta}$$

I computed this with explicit coordinates and computations learned from projective geometry:

$$ \left( \begin{pmatrix}a\cos\alpha\\a\sin\alpha\\1\end{pmatrix} \times \begin{pmatrix}b\cos\beta\\-b\sin\beta\\1\end{pmatrix} \right) \times \begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix} ab\sin\alpha\cos\beta + ab\cos\alpha\sin\beta \\ 0 \\ a\sin\alpha + b\sin\beta \end{pmatrix} $$

But there might be other ways to obtain that distance using more basic considerations. Now plug that expression into your inequality and you obtain

$$ \sin\alpha\left( \tfrac{ab\sin\alpha\cos\beta + ab\cos\alpha\sin\beta}{a\sin\alpha + b\sin\beta} -b\right) + \sin\beta\left( \tfrac{ab\sin\alpha\cos\beta + ab\cos\alpha\sin\beta}{a\sin\alpha + b\sin\beta} -a\right) \le 0 $$

Since $a\sin\alpha + b\sin\beta>0$ you can multiply everything by that denominator. You can also divide by $ab>0$. You obtain the following formula:

\begin{multline*} \sin^2\alpha\cos\beta + \sin\alpha\cos\alpha\sin\beta - \sin^2\alpha - \tfrac ba\sin\alpha\sin\beta \\ + \sin\alpha\sin\beta\cos\beta + \cos\alpha\sin^2\beta - \tfrac ab\sin\alpha\sin\beta - \sin^2\beta \le 0 \end{multline*}

You can regroup terms like this:

$$ \sin^2\alpha\left(\cos\beta-1\right) + \sin^2\beta\left(\cos\alpha-1\right) + \sin\alpha\sin\beta\left(\cos\alpha+\cos\beta-\tfrac ab-\tfrac ba\right) \le 0 $$

With $\cos\varphi\le1$ for all $\varphi$, you can show that each of the three summands is non-positive. The first two parentheses are easy, and for the last parenthesis you can use

$$\cos\alpha+\cos\beta-\tfrac ab-\tfrac ba\le 2-\tfrac ab-\tfrac ba = -\tfrac ab\left(1-\tfrac ba\right)^2\le 0$$

Thus the sum as a whole is non-positive.