Trigonometric inequality in an obtuse triangle

Question

Let $ABC$ be an obtuse triangle with $A$ the obtuse angle. I conjecture that the following inequality is true $$\sin B + \sin C \le |\tan A|.$$ Show that it holds or give a counterexample.

Answer 1

Since $B ,C , B+C < \frac{\pi}{2} $ we have $\frac{\cos B }{\cos (B +C )} >1$ and $\frac{\cos C }{\cos (B+C )} >1$ hence $$|\tan A|=\tan (B+C) =\frac{\cos B}{\cos (B+C)} \cdot \sin C + \frac{\cos C}{\cos (B+C)} \cdot \sin B >1\cdot \sin C +1\cdot \sin B =\sin C +\sin B .$$

Answer 2

Your conjecture seems correct. First, as $-\sin$ is convex in $[0,\frac{\pi}{2}]$, you have

$\sin B +\sin C \le 2\sin \frac{B+C}{2}.$

Let $\frac{B+C}{2}=\beta \in [0,\frac{\pi}{4})$. It remains to show that $$ 2\sin \beta \le \frac{\sin 2\beta} {\cos 2\beta}, $$ or $\cos 2\beta\le \cos \beta$. This follows from that $\cos(x)$ is decreasing in $[0,\frac{\pi}{2}]$.