Hmmm I'm surprised no one has posted a solution to this yet. Anyways here goes: We first prove a couple lemmas. Let $M(ABC)$ be the length of the longest side of triangle $ABC$. Lemma 1. Given a point P either inside or on a side of triangle ABC, $M(ABP) \le M(ABC)$. Suppose on the contrary that this is false; then WLOG $AP > M(ABC)$. But extend AP to hit BC at Q, and suppose R is on AB such that RQ || AC. We have $AP \le AQ \le AR+RQ = \frac{CQ \cdot AB}{BC}+\frac{BQ \cdot AC}{BC}\le \max(AC, BC) \le M(ABC)$, a contradiction. Hence lemma 1 is proven. Lemma 2. Given a point P on segment AB of any triangle ABC, we have $m(PAC) \le m(ABC)$. Let Q be the point on AC s.t. BC || PQ. By lemma 1, $M(PAC) \ge M(AQP)$. Then $m(PAC) = \frac{2[PAC]}{M(PAC)}\le \frac{2[PAC]}{M(AQP)}= \frac{2[ABC]}{M(ABC)}= m(ABC)$, where the second-to-last equality comes from $[PAC] = \frac{AP}{AB}\cdot [ABC]$ and $M(AQP) = \frac{AP}{AB}\cdot M(ABC)$. This proves lemma 2. Now consider the original problem. Case I: X lies in the interior of or on a side of ABC. $m(XAB)+m(XBC)+m(XCA) = \frac{2[XAB]}{M(XAB)}+\frac{2[XBC]}{M(XBC)}+\frac{2[XCA]}{M(XCA)}$ $\ge \frac{2[XAB]+2[XBC]+2[XCA]}{M(ABC)}= \frac{2[ABC]}{M(ABC)}= m(ABC)$, with the inequality resulting from lemma 1. Case II: One of the vertices of ABC lies in the interior of or on a side of the triangle formed by X and the other two vertices (WLOG A lies in XBC). Extend BA to meet XC at D. We have $m(XAB)+m(XBC)+m(XCA) \ge m(XBC) \ge m(BDC) \ge m(BAC)$ where the last two inequalities follow from lemma 2. Case III: A, B, C, and X form a convex quadrilateral (WLOG in that order). Let AC and BX meet at Y. Using lemma 2, we have $m(XAB)+m(XBC)+m(XCA) \ge m(AYB)+m(CYB)+m(XCA)$ $\ge m(AYB)+m(CYB) = m(AYB)+m(CYB)+m(AYC)$. However, as was shown in Case I, $m(AYB)+m(CYB)+m(AYC) \ge m(ABC)$. Therefore, the inequality holds in all three cases, and these are all the cases, so we are done.