Estonia — Open Contests

Let us first prove the second inequality. Consider the circle with centre $A$ that passes through the vertex $C$ and consider the extension of the side $CB$ past the vertex $B$ up to this circle (Fig. 5). As a chord lies inside the circle, we have $AD \le AC$. We also have $BE \le \max(BC, BA) = BC$ and $CF \le \max(CA, CB) = CB$. Since by the triangle inequality $\frac{1}{2}BC < \frac{1}{2}CA + \frac{1}{2}AB$, we obtain $AD + BE + CF \le 2BC + CA < \frac{3}{2}BC + \frac{3}{2}CA + \frac{1}{2}AB < \frac{3}{2}(BC + CA + AB)$.

Let us now prove the first inequality. If the triangle is not acute then $BE \ge BA$ and $CF \ge CA$ (Fig. 6). Since by the triangle inequality we have $\frac{1}{2}BC < \frac{1}{2}CA + \frac{1}{2}AB$, we obtain $AD + BE + CF \ge AD + BA + CA > BA + CA > \frac{1}{2}(BC + CA + AB)$. For an acute triangle we have $AD \ge AK$, $BE \ge BL$ and $CF \ge CM$, where $AK$, $BL$ and $CM$ are the altitudes of the triangle $ABC$. Let $H$ be the intersection point of the altitudes of the triangle $ABC$ (Fig. 7). We obtain $AD + BE + CF \ge AK + BL + CM > AH + BH + CH = \frac{1}{2}(BH + CH) + \frac{1}{2}(CH + AH) + \frac{1}{2}(AH + BH) > \frac{1}{2}(BC + CA + AB)$, where the last inequality follows from the triangle inequality.

Fig. 5

Fig. 6

Fig. 7

By the triangle inequality, $AB \le BD + AD$ and $AB \le AE + BE$, analogous inequalities hold for sides $BC$ and $CA$. As not all equality cases can hold simultaneously, adding these inequalities gives a strict inequality $2(BC + CA + AB) < BC + CA + AB + 2(AD + BE + CF)$. Collecting similar terms and dividing by 2 gives the first required inequality.

Similarly by the triangle inequality, $AD \le AB + BD$ and $AD \le CA + CD$; analogously for $BE$ and $CF$. Again, not all equality cases can hold simultaneously, whence adding these inequalities gives a strict inequality $2(AD + BE + CF) < 2(BC + CA + AB) + BC + CA + AB$. Collecting similar terms and dividing by 2 leads to the second required inequality.