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

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 $$\sin A < 1/2$$.

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

• (1) Commenting doesn't draw any attention to your question. (2) You should probably expect to wait far more than five minutes for people to give your question any consideration. Be patient. The Math.SE community is full of humans, not instant-answering robots. :)
– Blue
May 30, 2020 at 11:48
• @Blue thanks for suggestion it is now well noted in my mind. :) May 30, 2020 at 11:50

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.

• I think the "WLOG" assumption you've already made is sufficient, but I've noted it explicitly now in the answer. May 30, 2020 at 12:06

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.

• I've added something new, maybe that can help. May 30, 2020 at 12:12