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]
Best Answer
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.