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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?

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2 Answers 2

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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: $$ A=B=C=\frac\pi3 $$

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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 $A=B=C=\frac \pi 3$

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    $\begingroup$ Note that this has been asked and answered many times ... $\endgroup$
    – Martin R
    Commented Nov 20, 2021 at 8:34

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