Maximum value of $\sin(A/2)+\sin(B/2)+\sin(C/2)$? So I came across a question in my textbook:
In triangle ABC, if $A$,$B$,$C$ represent angles, then find the maximum value of $\sin(A/2)+\sin(B/2)+\sin(C/2)$?
So I have already tried and best and put my blood,sweat and tears into this question..But I'm not able to go solve further!
So here's my approach:
By Using $\sin(C)+\sin(D)$ and $A+B+C= \pi$;

*

*$2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+\cos(\frac{A+B}{2})$
Now,
Using $\cos(2A)$ formula i.e, $1-2\sin^2(A)
$

*$2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+1-2\sin^2(\frac{A+B}{4})$

*So I got quadratic in variable $\sin(\frac{A+B}{4})$

*$-2\sin^2(\frac{A+B}{4})+2\sin(\frac{A+B}{4})\cos(\frac{A-B}{4})+1$
But I dunno what to do After that
Can I solve this question using this method or I have to use a different approach!
BTW, the answer is 3/2
Edit:I have just finished my high school and preparing for entrance exam IIT-JEE,So please don't use hard terms to solve this question.
This solution is sent by my teacher, atleast make me understand this one
[https://i.stack.imgur.com/51pCB.png]
 A: Since $\sin x$ is concave on acute $x$, by Jensen's inequality the maximum is found at $A/2=B/2=C/2=\pi/6$, as $3\sin\pi/6=3/2$.
Edit: since the OP mentioned in a comment on @B.Goddard's answer that they know differentiation, here's another proof the equilateral case achieves a maximum:
Keep using $\frac{C}{2}=\frac{\pi}{2}-\frac{A+B}{2}$. To extemize $\sin\frac{A}{2}+\sin\frac{B}{2}+\cos\frac{A+B}{2}$ simultaneously solve$$\tfrac12\cos\tfrac{A}{2}-\tfrac12\cos\tfrac{C}{2}=0,\,\tfrac12\cos\tfrac{B}{2}-\tfrac12\cos\tfrac{C}{2}=0$$viz. $A=B=C$. I'll leave the reader to check it's a maximum by considering second derivatives.
A: You can do it with Lagrange multipliers.  Maximize $f=\sin x/2 + \sin y/2+\sin z/2$  under the constraint $g=x+y+z = \pi$.
Then
$$\nabla f = \langle \cos(x/2)/2, \cos(y/2)/2, \cos(z/2)/2 
\rangle =\lambda\langle 1,1,1 \rangle = \nabla g.$$
This shows that $x=y=z$ and the maximal triangle is equilateral.
A: Where you have stopped,
let $$z=-2\sin^2x+1+2\sin x\cos y$$
$$\iff2\sin^2x-2\sin x\cos y+z-1=0$$
As $\sin x$ is real, the discriminant must be $\ge0$
$\implies8(z-1)\le(-2\cos y)^2\le2^2$
$\implies8z\le4+8$
The equality occurs if $\cos^2y=1\iff\sin y=0$
and consequently $\sin x=\mp\dfrac{\cos y}2=\mp\dfrac12$
A: In a triangle ABC, $A+B+C=\pi$
$$f(x)=\sin(x/2) \implies f''(x)=-\frac{1}{4}\sin(x/2)<0, x\in[0,2\pi].$$ So by Jemsen's inequality $$\frac{f(A/2)+f(B/2)+f(C/2)}{3} \le f(\frac{A+B+C}{6}).$$
$$\implies \frac{\sin (A/2)+\sin(B/2)+\sin{C/2}}{3} \le \sin\frac{\pi}{6}.$$
$$\implies \sin(A/2)+\sin(B/2)+\sin(C/2) \le \frac{3}{2}$$
