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 $$\cot\measuredangle MAB + \cot\measuredangle MBC + \cot\measuredangle MCA \geq 3\sqrt{3}$$
 A: Assume that:
$$M=\alpha A+\beta B+\gamma C,\quad \alpha,\beta,\gamma\geq 0,\;\alpha+\beta+\gamma=1, $$
i.e. let $[\alpha,\beta,\gamma]$ be the barycentric coordinates of $M$. The line through $M$ and $A$ has equation $\gamma y-\beta z=0$ while the line through $A$ and $B$ has equation $z=0$. By corollary $18$ of Volonec,
$$\cot\widehat{MAB} = \cot A+(\cot A+\cot B)\,\frac{\beta}{\gamma}$$
hence:
$$\cot\widehat{MBC} = \cot B+(\cot B+\cot C)\,\frac{\gamma}{\alpha}$$
$$\cot\widehat{MCA} = \cot C+(\cot C+\cot A)\,\frac{\alpha}{\beta}$$
and the problem boils down to proving that:
$$ \left(1+\frac{\alpha}{\beta}+\frac{\beta}{\gamma}\right)\cot A+\left(1+\frac{\beta}{\gamma}+\frac{\gamma}{\alpha}\right)\cot B+\left(1+\frac{\gamma}{\alpha}+\frac{\alpha}{\beta}\right)\cot C \geq 3\sqrt{3}.$$
A: $\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= \\ $
$\displaystyle\sum_{cyc}\cot A+\sum_{cyc}\dfrac {\alpha} {\beta} \cdot \dfrac {\sin B} {\sin A\cdot \sin C} \ge \sqrt3 +3\dfrac 1 {\left(\sin A\cdot \sin B\cdot \sin C\right)^{\frac 1 3}}\ge \sqrt3 +2\sqrt3=3\sqrt3$
A: 
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
