If the angles of $\triangle ABC$ satisfy $\sin A+\sin B+\sin C\leq 1$, then show that one of them is more than $150^\circ$

Question

This is an Olympiad question.

In a $\triangle ABC$, where $\sin A+\sin B+\sin C\leq 1$, prove that one of the angles is more than $150^\circ$.

First of all I assumed that WLOG, $A\geq B \geq C$. Then I tried solving the problem using triangle inequality and Sine rule. From there, and from the given statement in the question, I was able to establish that $\sin A < \frac{1}{2}$.

I know that I just need to establish that angle $A$ is greater than $90^\circ$ or something like that.

Answer 1

You have shown that $\sin A < \frac12.$ Given that $0 < A < 180^\circ$ (according to the usual way of measuring angles of a triangle), this implies that either $A < 30^\circ$ or $A > 150^\circ.$

Now consider the implications of $A < 30^\circ$ along with the fact that $A + B + C = 180^\circ.$

You have already assumed $A > B > C$, so you can continue to use that assumption.

Answer 2

In an acute angled triangle, by Karamata inequality, $$\sin A+\sin B+\sin C \ge \sin (\pi/2)+\sin (\pi/2)+\sin 0=2$$ because $\sin x$ is concave, $(A,B,C)\prec (\pi/2,\pi/2,0)$.

Other ways to show the inequality: Question about sines of angles in an acute triangle

https://artofproblemsolving.com/community/c6h17035p117354

So, one of angle (let be A) is $\ge \pi/2$. If $A< 5\pi/6$, then $B+C> \pi/6$ and $\sin A> 1/2$

Because, $\cos C\le 1$ and $\cos B\le 1$ holds,

$$\sin B+\sin C\ge \sin B \cos C+\sin C \cos B=\sin(B+C)>1/2$$ which is contradiction.