<s>[INST] Let $M$ be an arbitrary point located inside the triangle $ABC$. Prove that $\cot\angle MAB + \cot\angle MBC + \cot\angle MCA \geq 3\sqrt{3}$ Let $M$ be an arbitrary point located inside the triangle $ABC$. Prove that $$\cot\measuredangle MAB + \cot\measuredangle MBC + \cot\measuredangle MCA \geq 3\sqrt{3}$$
 [/INST] $\cot A+\cot B+\cot C\ge \sqrt3 \ $ and $\ \sin A\cdot \sin B\cdot \sin C\le \dfrac {3\sqrt3} 8$
are well known inequalities
$(1+\dfrac {\alpha} {\beta} +\dfrac {\beta} {\gamma})\cot A+(1+\dfrac {\beta} {\gamma} +\dfrac {\gamma} {\alpha})\cot B+(1+\dfrac {\gamma} {\alpha} +\dfrac {\alpha} {\beta})\cot C= \\ $
$\displaystyle\sum_{cyc}\cot A+\sum_{cyc}\dfrac {\alpha} {\beta} \cdot \dfrac {\sin B} {\sin A\cdot \sin C} \ge \sqrt3 +3\dfrac 1 {\left(\sin A\cdot \sin B\cdot \sin C\right)^{\frac 1 3}}\ge \sqrt3 +2\sqrt3=3\sqrt3$
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