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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?

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  • $\begingroup$ Tip for MCQs: Most of the times triangle questions involving max/min values can be solved by assuming the triangle to be equilateral. $\endgroup$ Oct 25, 2014 at 15:52
  • $\begingroup$ Eg: min value of $\sum Tan$ $\endgroup$ Oct 25, 2014 at 15:53
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    $\begingroup$ See enotes.com/homework-help/… $\endgroup$ May 10, 2016 at 5:20
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    $\begingroup$ 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. $\endgroup$ May 10, 2016 at 5:21
  • $\begingroup$ I have edited my question,thanks for pointing @choco_addicted. $\endgroup$ May 10, 2016 at 5:26

5 Answers 5

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

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    $\begingroup$ Answer is wrong $\endgroup$ Dec 30, 2020 at 7:32
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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 $B=C$

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  • $\begingroup$ Hi. When we put $B=C$, we get the product as $2\sin\frac{2B}2=2\sin B$, which is maximum when $B=\frac\pi2$ but that's not what we want. What's my mistake? $\endgroup$
    – aarbee
    Feb 3 at 5:49
  • $\begingroup$ @aarbee My observation works for two angles, but the original expression has three. If there are two angles of the original three which are not equal, I can use this observation to show that it can't be a maximum. $\endgroup$ Feb 3 at 8:13
  • $\begingroup$ Are we saying that by going in circular way we can get $A=B=C?$ $\endgroup$
    – aarbee
    Feb 3 at 11:16
  • $\begingroup$ @aarbee If two of the three are different we can't possibly have a maximum, since by this observation we can increase the value, so the only possible maximum is with all three equal. If you choose the highest and lowest of the three every time and iterate, it is easy to see that the largest value decreases and the least value increases and the process converges (if you need to do that). $\endgroup$ Feb 3 at 20:39
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$$f(x,y,z)=\sin(x)+\sin(y)+\sin(z)$$ $$g(x,y,z)=x+y+z-\pi=0$$

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

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

hence

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

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

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

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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 $\sqrt{3}$. $\blacksquare$

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