Maximum value of $\sin A+\sin B+\sin C$?
What is the maximum value of $\sin A+\sin B+\sin C$ in a triangle $ABC$. My book says its $3\sqrt3/2$ but I have no idea how to prove it.
I can see that if $A=B=C=\frac\pi3$ then I get $\sin A+\sin B+\sin C=\frac{3\sqrt3}2$. And also according to WolframAlpha maximum is attained for $a=b=c$. But this does not give me any idea for the proof.
Can anyone help?
Solution 1:
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 $A=B=C=\pi/3$.
Solution 2:
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 $B=C$
Solution 3:
$$f(x,y,z)=\sin(x)+\sin(y)+\sin(z)$$ $$g(x,y,z)=x+y+z-\pi=0$$
$$\large\frac{\frac{\partial f}{\partial x}}{\frac{\partial g}{\partial x}}= \frac{\frac{\partial f}{\partial y}}{\frac{\partial g}{\partial y}}= \frac{\frac{\partial f}{\partial z}}{\frac{\partial g}{\partial z}}=k$$
$$\cos(x)=\cos(y)=\cos(z)$$
hence
$$f_{max}=\sin\left(\frac{\pi}{3}\right)+\sin\left(\frac{\pi}{3}\right)+\sin\left(\frac{\pi}{3}\right)=\frac{3\sqrt3}{2}$$
$$\sin(x)+\sin(y)+\sin(z)\le\frac{3\sqrt3}{2}$$
Solution 4:
The function $f(\alpha,\beta,\gamma):=\sin\alpha+\sin\beta+\sin\gamma$ assumes a maximum on the simplex $$S:=\bigl\{(\alpha,\beta,\gamma)\>\bigm|\>\alpha\geq0, \ \beta\geq0,\ \gamma\geq0,\ \alpha+\beta+\gamma=\pi\bigr\}\ .$$ On the other hand, if $\alpha>\beta\geq0$ one has $$\sin\alpha+\sin\beta=2\sin{\alpha+\beta\over2}\cos{\alpha-\beta\over2}<2\sin{\alpha+\beta\over2}\ ,$$ and $\bigl({\alpha+\beta\over2},{\alpha+\beta\over2},\gamma\bigr)\in S$.
This allows to conclude that ${\rm argmax}_S f=\bigl({\pi\over3},{\pi\over3},{\pi\over3}\bigr)$, so that$$f(\alpha,\beta,\gamma)\leq {3\sqrt{3}\over2}\qquad\bigl((\alpha,\beta,\gamma)\in S\bigr)\ .$$