Proving $\frac{1}{\sin(A/2)}+\frac{1}{\sin(B/2)}+\frac{1}{\sin(C/2)}\ge 6$, where $A$, $B$, $C$ are angles of a triangle

Question

If $A$, $B$, and $C$ are the angles of a triangle, then $$\frac{1}{\sin \left(\frac{A}{2}\right)}+\frac{1}{\sin\left(\frac{B}{2}\right)}+\frac{1}{\sin\left(\frac{C}{2}\right)}\ge 6$$

I have used multiple trigonometric identities, but the situation becomes complicated. I also thought about the Sine Law. To be honest, I don’t think these techniques are suitable. Any suggestions?

Answer 1

We have that

$$f(x)=\frac1{\sin x}$$

is convex then by Jensen's inequality

$$\frac{\frac{1}{\sin \left( \frac{A}{2} \right)}+\frac{1}{\sin \left( \frac{B}{2} \right)}+\frac{1}{\sin \left( \frac{C}{2} \right)}}3\ge \frac1{\sin\left(\frac{A+B+C}6\right)}$$

and then

$$\frac{1}{\sin \left( \frac{A}{2} \right)}+\frac{1}{\sin \left( \frac{B}{2} \right)}+\frac{1}{\sin \left( \frac{C}{2} \right)}\ge \frac3{\sin\left(\frac{A+B+C}6\right)}=6$$

Answer 2

HINT:

Consider $$f\left( x \right)={1}/{\sin \left( \frac{x}{2} \right)}\;,\quad 0<x<\pi $$

It easy to verify that ${f}''\left( x \right)>0$ and we have a convex function. Now Using Jensen’s inequality: $$\frac{1}{3}f\left( \frac{A}{2} \right)+\frac{1}{3}f\left( \frac{B}{2} \right)+\frac{1}{3}f\left( \frac{C}{2} \right)\ge f\left( \frac{1}{3}\frac{A}{2}+\frac{1}{3}\frac{B}{2}+\frac{1}{3}\frac{C}{2} \right)$$

Answer 3

Also, by AM-GM in the standard notation we obtain: $$\sum_{cyc}\frac{1}{\sin\frac{\alpha}{2}}\geq\frac{3}{\sqrt[3]{\prod\limits_{cyc}\sin\frac{\alpha}{2}}}=\frac{3}{\sqrt[3]{\frac{r}{4R}}}\geq\frac{3}{\sqrt[3]{\frac{1}{4\cdot2}}}=6.$$

Answer 4

We use $$\sin\left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{bc}}$$ etc then we get by AM-GM: $$\frac{1}{3}\left(\sqrt{\frac{bc}{(s-b)(s-c)}}+\sqrt{\frac{ac}{(s-a)(s-c)}}+\sqrt{\frac{ab}{(s-a)(s-b)}}\right)\geq \sqrt[3]{\frac{abc}{(s-a)(s-b)(s-c)}}\geq 2$$ if $$\frac{abc}{(s-a)(s-b)(s-c)}\geq 8$$

and this is $$abc\geq (-a+b+c)(a-b+c)(a+b-c)$$ using the Substitution

$$a=y+z,b=x+z,c=x+y$$

we get

$$(x+y)(x+z)(z+x)\geq 8xyz$$ and this is AM-GM! Note that $$s=\frac{a+b+c}{2}$$

Answer 5

Applying Lagrange Multipliers shows that $$ \sin(A/2) + \sin (B/2) + \sin (B/2)$$ is maximized when the triangle is equilateral in which case we have $$ \sin(A/2) + \sin (B/2) + \sin (B/2)=3/2$$

The Arithmetic Harmonic inequality implies $$\frac{1}{\sin \left( \frac{A}{2} \right)}+\frac{1}{\sin \left( \frac{B}{2} \right)}+\frac{1}{\sin \left( \frac{C}{2} \right)}\ge \frac9{\sin(A/2) +\sin(B/2)+\sin(C/2)}\ge \frac {9}{(3/2)}=6$$

Answer 6

A Pure Geometric Proof

Let $I$ be the incenter of $\triangle ABC$, $r$ be the radius of the incircle. Then $$\dfrac{r}{\sin\dfrac{A}{2}}+\dfrac{r}{\sin\dfrac{B}{2}}+\dfrac{r}{\sin\dfrac{C}{2}}=IA+IB+IC.\tag1$$

Now, let's apply Erdos-Mordell Inequality，which states that:

from a point $P$ inside a given $\triangle ABC$， the perpendiculars $PX, PY, PZ$ are drawn to its sides. Then

$$PA+PB+PC≥2(PX+PY+PZ).$$

Thus, we instantly have

$$IA+IB+IC\geq 2(r+r+r)=6r.\tag2$$ Combining $(1)$ and $(2)$, we readily obtain what we want to prove.