Maximum value of $\sin A+\sin B+\sin C$?

For $x\in[0,\pi]$, the function $f(x)=\sin(x)$ is concave, so by Jensen's inequality, we have $$ \frac{1}{3}f(A)+\frac{1}{3}f(B)+\frac{1}{3}f(C)\leq f\left[\frac{1}{3}(A+B+C)\right]=\sin(\pi/3)=\frac{\sqrt{3}}{2}. $$ Equality is achieved when $A=B=C=\pi/3$.

Here is a hint, which should get you most of the way there

Note that $\sin B+\sin C= 2\sin \frac {B+C}2 \cos \frac {B-C}2$

If $A$ is fixed then $B+C$ is fixed, and the product is greatest when $B=C$

$$f(x,y,z)=\sin(x)+\sin(y)+\sin(z)$$ $$g(x,y,z)=x+y+z-\pi=0$$

$$\large\frac{\frac{\partial f}{\partial x}}{\frac{\partial g}{\partial x}}= \frac{\frac{\partial f}{\partial y}}{\frac{\partial g}{\partial y}}= \frac{\frac{\partial f}{\partial z}}{\frac{\partial g}{\partial z}}=k$$

$$\cos(x)=\cos(y)=\cos(z)$$

hence

$$f_{max}=\sin\left(\frac{\pi}{3}\right)+\sin\left(\frac{\pi}{3}\right)+\sin\left(\frac{\pi}{3}\right)=\frac{3\sqrt3}{2}$$

$$\sin(x)+\sin(y)+\sin(z)\le\frac{3\sqrt3}{2}$$