Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.
I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a solution, but hit and trial. Can this conclusion be derived using maximum/mininum concept?
 A: Note that $\sin x$ is concave function when $0\leq x\leq \dfrac{\pi}{2}$. Then by Jensen's inequality on concave function, we have:
$$\frac{\sin \dfrac{A}{2}+\sin \dfrac{B}{2}+\sin \dfrac{C}{2}}{3}\leq \sin \dfrac{A+B+C}{2\cdot 3}=\sin\dfrac{\pi}{6}=\frac{1}{2}$$
Now by AM-GM inequality $$\frac{\sin \dfrac{A}{2}+\sin \dfrac{B}{2}+\sin \dfrac{C}{2}}{3}\geq \Big(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\Big)^{\dfrac{1}{3}}$$
Hence $$\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$$
Equality holds when $A=B=C=\dfrac{\pi}{3}$
A: This question is the same as asking: when $\alpha+\beta+\gamma=\frac\pi2$, what is the maximum of $\sin(\alpha)\sin(\beta)\sin(\gamma)$?
We wish to find $\alpha,\beta,\gamma$ so that for each $\delta\alpha,\delta\beta,\delta\gamma$ so that
$$
\delta\alpha+\delta\beta+\delta\gamma=0\tag{1}
$$
we also have
$$
\begin{align}
0
&=\delta\sin(\alpha)\sin(\beta)\sin(\gamma)\\
&=\cos(\alpha)\sin(\beta)\sin(\gamma)\,\delta\alpha+\sin(\alpha)\cos(\beta)\sin(\gamma)\,\delta\beta+\sin(\alpha)\sin(\beta)\cos(\gamma)\,\delta\gamma\tag{2}
\end{align}
$$
Since each $\delta\alpha,\delta\beta,\delta\gamma$ that is perpendicular to $(1,1,1)$, it must also be perpendicular to $(\cot(\alpha),\cot(\beta),\cot(\gamma))$, we must have that $\cot(\alpha)=\cot(\beta)=\cot(\gamma)$. Thus, $\alpha=\beta=\gamma=\frac\pi6$ and
$$
\sin(\alpha)\sin(\beta)\sin(\gamma)=\frac18\tag{3}
$$
A: Like my answers in
Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $
Trigonometric Inequality $\sin{A}+\sin{B}-\cos{C}\le\frac32$
In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$
let $S=\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$
$2S=\left(\cos\dfrac{A-B}2-\cos\dfrac{A+B}2\right)\sin\dfrac C2$
As $\dfrac{A+B}2=\dfrac\pi2-\dfrac C2,$
$2S=\left(\cos\dfrac{A-B}2-\sin\dfrac C2\right)\sin\dfrac C2$
$\iff\sin^2\dfrac C2-\sin\dfrac C2\cos\dfrac{A-B}2+2S=0$
As $\sin\dfrac C2$ is real, the discriminant must be $\ge0$
i.e., $$\cos^2\dfrac{A-B}2-8S\ge0\iff8S\le\cos^2\dfrac{A-B}2\le1$$
So, $8S\le1\implies S\le\frac{1}{8}$
the equality occurs if $\cos\dfrac{A-B}2=1\iff\cos(A-B)=1\iff A=B$ as $0<A,B<\pi$
and hence $\sin\dfrac C2=\dfrac12\implies C=\dfrac\pi3$ as $0<C<\pi$
$\implies A=B=\dfrac{A+B}2=\dfrac\pi3=C$
A: $$\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}=\frac{r}{4R}\leq\frac{1}{8}.$$
Another way.
Since $f(x)=\ln\sin\frac{x}{2}$ is a concave function, by Jensen and AM-GM we obtain:
$$\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\leq\sin^2\frac{\frac{\alpha}{2}+\frac{\beta}{2}}{2}\sin\frac{\gamma}{2}=$$
$$=\frac{1}{2}\left(1-\sin\frac{\gamma}{2}\right)\sin\frac{\gamma}{2}\leq\frac{1}{2}\left(\frac{1-\sin\frac{\gamma}{2}+\sin\frac{\gamma}{2}}{2}\right)^2=\frac{1}{8}.$$
Done!
A: Let us consider the function $f(\theta)=\log\sin\theta$ on the interval $\theta\in\left(0,\frac{\pi}{2}\right)$. It is a concave function
since $f''(\theta)=-\frac{1}{\sin^2\theta}<0$, hence by Jensen's inequality
$$ f(a)+f(b)+f(c) \leq 3\,f\left(\frac{a+b+c}{3}\right) $$
holds for any $a,b,c\in\left(0,\frac{\pi}{2}\right)$. By considering $a=\frac{A}{2},b=\frac{B}{2},c=\frac{C}{2}$ and exponentiating both sides of the previous inequality turns into
$$ \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\leq \sin^3\left(\frac{A+B+C}{6}\right) = \frac{1}{8}.$$
A: Let $a, b, c$ be the sides of a triangle with given angles $A, B, C$.
Recall for any $3$ positive numbers $a, b, c$, it can form the sides of a non-degenerate triangle if and only if we can find $3$ positive numbers
$u,v,w$ such that
$$a = v + w,\quad b = u + w,\quad c = u + v$$
This is known as Ravi substitution for a triangle.
In terms of them, we have
$$
\sin^2\frac{A}{2} 
= \frac12( 1 - \cos A) = \frac12\left(1 - \frac{b^2+c^2-a^2}{2bc}\right)
= \frac{a^2 - (b-c)^2}{4bc} = \frac{vw}{bc}
$$
By a similar argument, we have
$$\sin^2\frac{B}{2}  = \frac{uw}{ac}\quad\text{ and }\quad\sin^2\frac{C}{2} = \frac{uv}{ab}$$
Multiply these 3 relations together, taking square root and apply AM $\ge$ GM to the factors in denominator, we obtain:
$$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}
= \frac{uvw}{abc}
= \frac{uvw}{(v+w)(u+w)(u+v)}
\le \frac{uvw}{(2\sqrt{vw})(2\sqrt{uw})(2\sqrt{uv})} = \frac18$$
