<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] 
Let $A_1,B_1,C_1$  be the intersection points of $AM,BM,CM$ with $BC,CA,AB$ respectively, and let $D,E,F$  be the feet of perpendiculars from $M$ to $BC,CA,AB$ respectively.Then we have 
\begin{align*}
\cot \angle MAB+\cot \angle MBC+\cot \angle MCA&=\dfrac {FA} {FM}+\dfrac {BD} {MD}+\dfrac {CE} {ME} \\
&=\sqrt{\dfrac {MA^2} {FM^2}-1}+\sqrt{\dfrac {MB^2} {MD^2}-1}+\sqrt{\dfrac {MC^2} {ME^2}-1}\\
&\ge \sqrt{\dfrac {MA^2} {MC_1^2}-1}+\sqrt{\dfrac {MB^2} {MA_1^2}-1}+\sqrt{\dfrac {MC^2} {MB_1^2}-1} 
\end{align*}
Using Van Aubel theorem we have
$$\dfrac {MA} {MA_1}=\dfrac {AC_1} {C_1B}+\dfrac {AB_1} {B_1C}$$
$$\dfrac {MB} {MB_1}=\dfrac {BA_1} {A_1C}+\dfrac {BC_1} {C_1A}$$
$$\dfrac {MC} {MC_1}=\dfrac {CB_1} {B_1A}+\dfrac {CA_1} {A_1B}$$
Multiplying these and applying AmGm inequality we get
$$AM\cdot BM\cdot CM\ge 8MA_1 \cdot MB_1\cdot MC_1$$
Let $\dfrac {MA} {MC_1}=a,\dfrac {MB} {MA_1}=b,\dfrac {MC} {MB_1}=c $ so ti is sufficies to prove
$$\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$$ where
$$a\ge1,b\ge1,c\ge1,abc\ge8$$
This one is an immediate consequence of Jensen inequality
</s>