What is wrong with my solution of the maximum value of $\displaystyle\sin \frac {A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2}$ in a triangle ABC?
I am NOT after the answer.
I know that $\displaystyle \sin \frac {A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \leq 1/8 $
And I also know that arithmetic mean is greater than equal to the geometric mean.
$\displaystyle \sin \frac {A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} \geq 3[{\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} }]^{1/3} $
$\displaystyle \sin \frac {A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} \geq 3/2 $
but this is wrong. Right is $ \sin \frac {A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} \leq 3/2 $
I am a high school student.
Best Answer
Let $a,\,g$ respectively denote the half-angles' sines' arithmetic and geometric means. You know $g\le\frac12$ and $a\ge g$, but that doesn't imply $a\ge\frac12$, and (as you've clearly read somewhere) we can in fact prove $a\le\frac12$.
Let's first note an equaliteral triangle obtains $a=g=\sin\frac{\pi}{6}=\frac12$, and now let's see if we can prove $\sum_{i=1}^3\sin\frac{A_i}{2}$ cannot exceed this with $A_1:=A,\,A_2:=B,\,A_3:=C$. Since $0\lt\frac{A_i}{2}\lt\frac{\pi}{2}\implies\sin^{\prime\prime}\frac{A_i}{2}=-\sin\frac{A_i}{2}<0$, it suffices to use Jensen's inequality for concave functions (Eq. (2) here).