In $\triangle ABC$, prove: $\frac{\sin^2 \frac A2}{\sin B \sin C}+\frac{\sin^2 \frac B2}{\sin A\sin C}+\frac{\sin^2 \frac C2}{\sin A\sin B} \ge 1$

Question

In a $\triangle ABC $, prove : $$\frac{\sin^2 \dfrac{A}{2}}{\sin B \sin C} + \frac{\sin^2 \dfrac{B}{2}}{\sin A\sin C} +\frac{\sin^2 \dfrac{C}{2}}{\sin A\sin B} \geq 1 $$ My approach :

$$2\left(\frac{\sin^2 \dfrac{A}{2}}{\sin B \sin C} + \frac{\sin^2 \dfrac{B}{2}}{\sin A\sin C} +\frac{\sin^2 \dfrac{C}{2}}{\sin A\sin B}\right) \geq 2 $$

$$\frac{1-\cos A}{\sin B \sin C} + \frac{1-\cos B}{\sin A\sin C} +\frac{1-\cos C}{\sin A\sin B} \geq 2 $$

Using $-\cos A = \cos (B+C) = \cos B \cos C -\sin B \sin C $

$$\frac{1-\cos A }{\sin B \sin C } = \frac{1}{\sin B \sin C} + \frac{ \cos B \cos C -\sin B \sin C}{\sin B \sin C} = \frac{1}{\sin B \sin C}+\cot B \ \cot C - 1 $$

$$ \frac{1}{\sin B \sin C} +\cot B \ \cot C - 1 + \frac{1}{\sin A \sin C} +\cot A \ \cot C - 1 + \frac{1}{\sin A \sin B} +\cot A \ \cot B - 1 \geq 2 $$

Using : $ A+B+C = \pi , \cot B \ \cot C +\cot A \ \cot C +\cot A \ \cot B = 1 $

$$ \frac{1}{\sin B \sin C} + \frac{1}{\sin A \sin C} + \frac{1}{\sin A \sin B} \geq 4 $$ It is equivalent to :

$$\frac{\sin A + \sin B + \sin C }{\sin A \sin B \sin C } \geq 4 $$

How do I proceed from here?

Answer 1

We have that by AM-GM

$$\frac{\sin A + \sin B + \sin C }{\sin A \sin B \sin C } \geq \frac{3^3 }{(\sin A + \sin B + \sin C)^2 } \stackrel{?}\ge 4$$

and the latter inequality is true indeed it is equivalent to

$$ \left(\frac{\sin A + \sin B + \sin C}3\right)^2\le \frac 34 \iff \frac{\sin A + \sin B + \sin C}3\le \frac{\sqrt 3}2$$

which is true by convexity.

Answer 2

Use $\sin C=\sin(A+B)$ and $\sin A+\sin B=2\sin \frac{A+B}2\cos\frac{A-B}2$, we get

$$\begin{align}\frac{\sin A + \sin B + \sin C }{\sin A \sin B \sin C } \geq 4&\Longleftrightarrow \frac{2\sin \frac{A+B}2\cos\frac{A-B}2 + 2\sin \frac{A+B}2\cos\frac{A+B}2 }{\sin A \sin B \cdot 2\sin \frac{A+B}2\cos\frac{A+B}2 } \geq 4 \\ \\ &\Longleftrightarrow\frac{\cos\frac{A-B}2+\cos\frac{A+B}2}{\sin A \sin B \cos\frac{A+B}2}\ge4 \\ \\ &\Longleftrightarrow\frac{2\cos\frac{A}2\cos\frac{B}2}{2\sin \frac A2 \cos\frac A2 \cdot2\sin \frac B2 \cos\frac B2\cdot\sin\left(\frac\pi2-\frac{A+B}2\right)}\ge4\\ \\ &\Longleftrightarrow \sin \frac A2\sin \frac B2\sin \frac C2\le\frac18\end{align}$$

This is true, because $$\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}= \frac{1}{2}\sin \frac{A}{2} \left(\cos \frac{B-C }{2}-\sin \frac{A}{2}\right)\le \frac{1}{2}\sin \frac{A}{2} \left(1-\sin \frac{A}{2}\right)\le \frac{1}{8} $$

Answer 3

Another way.

In the standard notation by C-S we obtain: $$\sum_{cyc}\frac{\sin^2\frac{\alpha}{2}}{\sin\beta\sin\gamma}=\sum_{cyc}\frac{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}}{\frac{2S}{ac}\cdot\frac{2S}{ab}}=\sum_{cyc}\frac{a^2(a+b-c)(a+c-b)}{(a+b+c)\prod\limits_{cyc}(a+b-c)}=$$ $$=\frac{1}{a+b+c}\sum_{cyc}\frac{a^2}{b+c-a}\geq\frac{1}{a+b+c}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(b+c-a)}=1.$$