Let $\triangle ABC$, $M\in Int(\triangle ABC)$. Let $MD\perp BC$, $ME\perp AC$ and $MF\perp AB$, $D\in BC$, $E\in AC$, $F\in AB$. Prove that $$MA+MB+MC\geq 2\cdot(MD+ME+MF).$$
My only idea is to use the fact that $MA+MD>AD$ and the similars and I would need to prove that $$AD+BE+CF\geq 3\cdot (MD+ME+MF)$$ but I did not manage to do that.
Let us use Barrow's inequality. with notations mixing yours and those of this Wikipedia article:
$$MA+MB+MC \ge 2(MU+MV+MW) \ge 2(MD+ME+MF)$$
because the altitudes lengths, in each triangle $MAB$, $MBC$, $MCA$ is shorter than the angle bissector lengths.
The case of equality is clearly when $ABC$ is equilateral.