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- Feb 14, 2012
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For any triangle $ABC$, prove that
$\cos \dfrac{A}{2} \cot \dfrac{A}{2}+\cos \dfrac{B}{2} \cot \dfrac{B}{2}+\cos \dfrac{C}{2} \cot \dfrac{C}{2} \ge \dfrac{\sqrt{3}}{2} \left( \cot \dfrac{A}{2}+\cot \dfrac{B}{2}+\cot \dfrac{C}{2} \right)$
$\cos \dfrac{A}{2} \cot \dfrac{A}{2}+\cos \dfrac{B}{2} \cot \dfrac{B}{2}+\cos \dfrac{C}{2} \cot \dfrac{C}{2} \ge \dfrac{\sqrt{3}}{2} \left( \cot \dfrac{A}{2}+\cot \dfrac{B}{2}+\cot \dfrac{C}{2} \right)$