# 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?

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.

• You know that the maximum for the geometric mean is $\dfrac{1}{2}$. But, consider the triangle where $A$ is very close to $\pi$ and $B$ and $C$ are both very close to zero. The geometric mean is very close to zero, but the arithmetic mean is very close to $1$. All you know is that the upper bound is $\ge \dfrac{3}{2}$. But, your argument does not tell you what that upper bound might be. It happens to also be $\dfrac{3}{2}$, but that is coincidence. Your argument tells you that $\dfrac{3}{2}$ is a lower bound for the upper bound of the arithmetic mean. Commented May 6, 2019 at 15:24
• @InterstellarProbe geometric mean would be 1/2 not 1/8. Commented May 6, 2019 at 15:26
• You are correct. But, my point still stands. Commented May 6, 2019 at 15:27
• @InterstellarProbe I am a high school student. I newer learned 'bounds'. Google search and wiki doesn't help me. It will be great if you could give me some source where I can study bound. Also 1/2 is still greater than 0 so what is the problem?? Commented May 6, 2019 at 15:33
• en.wikipedia.org/wiki/Upper_and_lower_bounds Commented May 6, 2019 at 16:13

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