Question:
Studying some geometric inequalities about arbitrary points, I thought of the following conjecture:
Define triangle $ABC$ and let $M$ be an arbitrary point inside triangle $ABC$. Let $MD \perp BC$ (with $D \in BC$), $ME \perp AC$ (with $E \in AC$), $MF \perp AB$ (with $F \in AB$). Then the following inequality holds:
$$\frac{MB \cdot MC}{MD} + \frac{MC \cdot MA}{ME} + \frac{MA \cdot MB}{MF} \geq 2(MA+MB+MC)$$
Attempt:
My attempt to prove this inequality is to show that $$\frac{MB \cdot MC}{MD} \geq MB+MC$$ but this idea seems wrong, considering that the inequality is equivalent to: $$\frac{1}{MD} \geq \frac{1}{MB}+\frac{1}{MC}$$ so to prove that $\sin \angle MBC + \sin \angle MCB \leq 1$, that cannot be true.
In what way can we prove (or disprove) this inequality?
Maybe the Erdos-Mordell inequality may have some application here: $$\frac{MA+MB+MC}{ME+MD+MF} \geq 2$$
This is indeed the Erdos-Mordell inequality in disguise.
Draw through the vertices $A,B,C$ the lines perpendicular to $MA,MB,MC$, respectively. Let the points of the pairwise intersection of these lines be $D',E',F'$ (see figure).
It remains to prove $$MD'=\frac{MB\cdot MC}{MD}$$
and similarly for $ME',MF'$. This proof I left to you as an exercise.