MHB Trigonometric Inequality Challenge

[anemone]

For any triangle $ABC$, prove that

$\cos \dfrac{A}{2} \cot \dfrac{A}{2}+\cos \dfrac{B}{2} \cot \dfrac{B}{2}+\cos \dfrac{C}{2} \cot \dfrac{C}{2} \ge \dfrac{\sqrt{3}}{2} \left( \cot \dfrac{A}{2}+\cot \dfrac{B}{2}+\cot \dfrac{C}{2} \right)$

 

[anemone]

For any triangle $ABC$, prove that

$\cos \dfrac{A}{2} \cot \dfrac{A}{2}+\cos \dfrac{B}{2} \cot \dfrac{B}{2}+\cos \dfrac{C}{2} \cot \dfrac{C}{2} \ge \dfrac{\sqrt{3}}{2} \left( \cot \dfrac{A}{2}+\cot \dfrac{B}{2}+\cot \dfrac{C}{2} \right)$

Soluion by David E. Narvaez, Panama:

Spoiler

From Jensen's inequality we have that

$\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2}\ge\sqrt{3}$ and $\sin\dfrac{A}{2}\sin\dfrac{B}{2}+\sin\dfrac{B}{2}\sin\dfrac{C}{2}+\sin\dfrac{C}{2}\sin\dfrac{A}{2} \ge\dfrac{3}{4}$ thus

$\displaystyle \dfrac{3}{2}\left( \sum_{cyc} \tan \dfrac{A}{2} \right)\left( \sum_{cyc} \sin \dfrac{B}{2} \sin \dfrac{C}{2} \right) \ge \dfrac{\sqrt{3}}{2}$

Let us assume, without loss of generality, that $A\ge B \ge C$. Then $\left( \tan \dfrac{A}{2} +\tan \dfrac{B}{2} \right) \ge \left( \tan \dfrac{A}{2} +\tan \dfrac{C}{2} \right) \ge \left( \tan \dfrac{B}{2} +\tan \dfrac{C}{2} \right)$ and

$\sin \dfrac{A}{2}\sin \dfrac{B}{2} \ge \sin \dfrac{C}{2} \sin \dfrac{A}{2} \ge \sin\dfrac{B}{2}\sin \dfrac{C}{2}$ and by Chebychev's inequality, we get

$\displaystyle \sum_{cyc} \left(\tan \dfrac{B}{2}+\tan \dfrac{C}{2} \right)\sin \dfrac{B}{2} \sin \dfrac{C}{2} \ge \dfrac{1}{3} \left(\sum_{cyc} \left(\tan \dfrac{B}{2}+\tan \dfrac{C}{2} \right) \right) \left( \sum_{cyc} \sin \dfrac{B}{2} \sin \dfrac{C}{2} \right) \ge \dfrac{\sqrt{3}}{2}$

but

$\begin{align*}

\left(\tan \dfrac{B}{2}+\tan \dfrac{C}{2} \right)\sin \dfrac{B}{2} \sin \dfrac{C}{2}&=\left(\dfrac{\sin \dfrac{B}{2} \cos \dfrac{C}{2}+\sin \dfrac{C}{2} \cos \dfrac{B}{2}}{\cos \dfrac{B}{2}\cos \dfrac{C}{2}} \right)\sin \dfrac{B}{2} \sin \dfrac{C}{2}\\&=\sin \dfrac{B+C}{2}\tan\dfrac{B}{2}\tan\dfrac{C}{2} \\&= \cos \dfrac{A}{2} \tan \dfrac{B}{2} \tan \dfrac{C}{2}\end{align*}$

and replacing this and similar identities for every term in the left hand side of our last inequality we have

$\displaystyle \sum_{cyc} \cos \dfrac{A}{2}\tan \dfrac{B}{2}\tan \dfrac{C}{2} \ge \dfrac{\sqrt{3}}{2}$

Multiplying this inequality by $\cot \dfrac{A}{2}\cot \dfrac{B}{2}\cot\dfrac{C}{2}=\cot \dfrac{A}{2}+\cot \dfrac{B}{2}+\cot\dfrac{C}{2}$ we get

$\cos \dfrac{A}{2} \cot \dfrac{A}{2}+\cos \dfrac{B}{2} \cot \dfrac{B}{2}+\cos \dfrac{C}{2} \cot \dfrac{C}{2} \ge \dfrac{\sqrt{3}}{2} \left( \cot \dfrac{A}{2}+\cot \dfrac{B}{2}+\cot \dfrac{C}{2} \right)$ and we are done.