How to solve this inequality.

Question

Let $\alpha$, $\beta$, $\gamma$ be the angles of a triangle. Show that $\sin\frac{\alpha}{2}.\sin\frac{\beta}{2}.\sin\frac{\gamma}{2}<{\frac{1}{4}}$

Answer 1

HINT:

$$2\sin\frac\alpha2\sin\frac\beta2=\cos\frac{\alpha-\beta}2-\cos\frac{\alpha+\beta}2$$

and $$\cos\frac{\alpha+\beta}2=\cos\frac{\pi-\gamma}2=\sin\frac\gamma2$$

Now use this method

Answer 2

As $0<\dfrac\alpha2<\dfrac\pi2,\sin\dfrac\alpha2>0$ etc.,

using A.M. $\ge$ G.M

$$\frac{\sin\dfrac\alpha2+\sin\dfrac\beta2+\sin\dfrac\gamma2}3\ge\sqrt[3]{\sin\dfrac\alpha2\cdot\sin\dfrac\beta2\cdot\sin\dfrac\gamma2} $$

Again as sine is concave in $[0,\pi]$ using Jensen's Inequality

$$\frac{\sin\dfrac\alpha2+\sin\dfrac\beta2+\sin\dfrac\gamma2}3\le\sin\left(\dfrac{\dfrac\alpha2+\dfrac\beta2+\dfrac\gamma2}3\right)=\sin\frac\pi6=\frac12$$