9
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Tip for MCQs: Most of the times triangle questions involving max/min values can be solved by assuming the triangle to be equilateral. $\endgroup$ – Harshal Gajjar Oct 25 '14 at 15:52
  • $\begingroup$ Eg: min value of $\sum Tan$ $\endgroup$ – Harshal Gajjar Oct 25 '14 at 15:53
  • 1
    $\begingroup$ See enotes.com/homework-help/… $\endgroup$ – lab bhattacharjee May 10 '16 at 5:20
  • 2
    $\begingroup$ 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. $\endgroup$ – choco_addicted May 10 '16 at 5:21
  • $\begingroup$ I have edited my question,thanks for pointing @choco_addicted. $\endgroup$ – Vinod Kumar Punia May 10 '16 at 5:26
7
$\begingroup$

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 $B=C$

$\endgroup$
19
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

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

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

hence

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.