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?

I'm sure there's a different way to approach this question, but here's one way using the graph of $\sin x $:

Consider $3$ points on the graph for $x\in(0,\pi)$. They are $(A,\sin A)$, $(B,\sin B)$, and $(C,\sin C)$ as shown. $A,B,C$ are such that $A+B+C=\pi$.
(Ignore the fact that $A,B$ and $C$ are angles of a triangle, they are just some values of $x$ that satisfy the condition $A+B+C=\pi$)
$A, B$ and $C$ are plotted on the graph of $y=\sin x$ and are joined to form a triangle, as shown.
Consider the centroid of the triangle, $G$, given by:
$$G=\left(\dfrac{A+B+C}{3}, \dfrac{\sin A+\sin B+\sin C}{3}\right)$$
Draw a line $PG$ as shown at $x=\dfrac{A+B+C}{3}$ (or $\dfrac{\pi}{3}$). This line intersects the curve at the point $P$ given by:
$$P=\left(\dfrac{A+B+C}{3}, \sin \left(\dfrac {A+B+C}{3}\right)\right)$$
From the figure, it is evident that the $y$-value of $P$ $>$ the $y$-value of $G$.
So we obtain the inequality $$\sin \left(\dfrac {A+B+C}{3}\right) >\dfrac{\sin A+\sin B+\sin C}{3}$$
Note that, for the case where $A=B=C$, we can deduce that $A=B=C= \dfrac{\pi}{3}$ and therefore $A,B$ and $C$ will coincide with point $P$ on the curve. For this particular case, we can deduce that $$\sin \left(\dfrac {A+B+C}{3}\right) =\dfrac{\sin A+\sin B+\sin C}{3}$$
Combining both the inequalities obtained, we get the desired result:
$$\sin \left(\dfrac {A+B+C}{3}\right) \geq \dfrac{\sin A+\sin B+\sin C}{3}$$
or
$$3 \sin \left(\dfrac {A+B+C}{3}\right) \geq \sin A+\sin B+\sin C$$
Credit for the answer: "Play with Graphs" by Amit Agarwal.
