<s>[INST] 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?
 [/INST] Note that
$$\sin A+ \sin B=2\sin \frac{A+B}2\cos\frac{{A-B}}2$$
Thus for any fixed $C$ the first 2 terms obtains a maximum for $A=B$. Similarly if we fix $A$ we get that there is a maximum when $B=C$. It follows that the maximum is obtained when:
$$
A=B=C=\frac\pi3
$$
</s>