Find the maximum of $\sin(A)+\sin(B)+\sin(C)$ for $\triangle ABC$. (Without Jensen Inequality) 
Find the maximum of $\sin(A)+\sin(B)+\sin(C)$ for $\triangle ABC$. (Without Jensen Inequality)

Proof of Jensen Inequality:
\begin{align}
&\text{let } f(x)=\sin x. \\
\ \\
\Rightarrow & f(A)+f(B)+f(C) \\ 
& =3 \bigg( \frac 1 3 f(A) + \frac 1 3 f(B)+ \frac 1 3 f(C) \bigg) \\
& \leq 3 \Bigg( f \bigg( \frac 1 3 A + \frac 1 3 B + \frac 1 3 C \bigg) \Bigg) \\
& = 3\Bigg(f\bigg( \frac {A+B+C} 3 \bigg) \Bigg) \\
& = 3\big(f(60)\big) & (\because A+B+C=180) \\
&=3\sin 60 = 3 \cdot \frac {\sqrt{3}} 2 = \frac {3\sqrt{3}}{2}.
\end{align}
I just wondered if there is another precalculus solution to this. Is there another solution to this?

Your solution is good; if you want another solution, use $C=\pi-A-B$ and then you look for the maximum value of
$$F=\sin (A)+\sin (B)+\sin (A+B)$$ Compute the partial derivatives
$$\frac{\partial F}{\partial A}=\cos (A)+\cos (A+B)=0\qquad \qquad \frac{\partial F}{\partial B}=\cos (B)+\cos (A+B)=0$$ Then, subtracting, $\cos(A)=\cos(B)$ and $A=B$ and then
$$\frac{\partial F}{\partial A}=\cos (A)+\cos (2A)=0$$
Use the double angle formula $$2\cos^2 (A)+\cos (A)-1=0$$ Solve the quadratic in $\cos(A)$ which gives $\cos(A)=\frac 12$ and then $A=B=C=\frac \pi 3$
