# Question about a geometric inequality

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: $$\frac{MA+MB+MC}{ME+MD+MF} \geq 2$$

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.