# 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?

• Tip for MCQs: Most of the times triangle questions involving max/min values can be solved by assuming the triangle to be equilateral. – Harshal Gajjar Oct 25 '14 at 15:52
• Eg: min value of $\sum Tan$ – Harshal Gajjar Oct 25 '14 at 15:53
• – lab bhattacharjee May 10 '16 at 5:20
• If $A=B=C=\frac{\pi}{2}$, then $\sin A + \sin B + \sin C= 3 > \frac{3\sqrt{3}}{2}$. I think the condition "$A$, $B$, and $C$ are angles of triangle $ABC$" should be added. – choco_addicted May 10 '16 at 5:21
• I have edited my question,thanks for pointing @choco_addicted. – Vinod Kumar Punia May 10 '16 at 5:26

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 $B=C$

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 $A=B=C=\pi/3$.

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$$f(\alpha,\beta,\gamma)\leq {3\sqrt{3}\over2}\qquad\bigl((\alpha,\beta,\gamma)\in S\bigr)\ .$$

Observe that $\sin A+\sin B=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})\leq2\sin(\frac{A+B}{2})=2\cos\frac{C}{2}$

so it it sufficient to show that $\sin C+2\cos \frac{C}{2}\leq \frac{3\sqrt{3}}{2}$

In fact assume $t=\cos\frac{C}{2}$, above becomes $2t\sqrt{1-t^2}+2t$, so finally we only need to estimate the upper boundary of $t\sqrt{1-t^2}+t$, where $t\in (0,1)$, while there are many ways to calculate for example using basic calculus, the boundary is $\sqrt{3}$. $\blacksquare$

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

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

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

hence

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

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