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

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Alternative solution.

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.