What is wrong with my solution of maximum value of $ \sin \frac {A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} $ in a triangle ABC?

Question

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.

Answer 1

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).

Answer 2

For $\alpha=\beta=\gamma$ we get a value $\frac{3}{2}.$

We'll prove that it's a maximal value.

Indeed, in the standard notation by AM-GM we obtain: $$\sum_{cyc}\sin\frac{\alpha}{2}=\sum_{cyc}\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}}=\frac{1}{2}\sum_{cyc}\sqrt{\frac{(a+b-c)(a+c-b)}{bc}}\leq$$ $$\leq\frac{1}{4}\sum_{cyc}\left(\frac{a+b-c}{b}+\frac{a+c-b}{c}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{b+c-a}{c}+\frac{a+c-b}{c}\right)=\frac{3}{2}.$$