In a triangle, prove that $ \sin A + \sin B + \sin C \leq 3 \sin \left(\frac{A+B+C}{3}\right) $ 
Prove that for any $\Delta ABC$ we have the following inequality:
$$ \sin A + \sin B + \sin C \le 3 \sin \left(\frac{A+B+C}{3}\right) $$

Could you use AM-GM to prove that?

We know in a triangle$ ABC, 0 \lt A,B,C \lt \pi$. So $sin A , sinB, sinC \gt 0$. Moreover in $[0,\pi],$ so $(sinx)''=-sinx \lt 0$, which is concave downwards. Using Jensen's inequality for concave downwards, we have:$
sin(\frac {A+B+C} 3)\ge \frac {sinA+sinB+sinC} 3$ or $sinA +sinB+sinC \le 3\sqrt 3 /2$
