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

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.
