Once I tried to find a solution which is substantially different from Kalva's ( http://www.kalva.demon.co.uk/imo/isoln/isoln811.html ). It was hard, but finally I found it. In the following, we will assume that the points D, E, F lie inside the segments BC, CA, AB; if they are not, then the proof needs a few adjustments (just use directed segments). We have $\frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF}=\frac{BD+DC}{PD}+\frac{CE+EA}{PE}+\frac{AF+FB}{PF}$ $=\frac{BD}{PD}+\frac{DC}{PD}+\frac{CE}{PE}+\frac{EA}{PE}+\frac{AF}{PF}+\frac{FB}{PF}$ $=\left(\frac{AF}{PF}+\frac{EA}{PE}\right)+\left(\frac{BD}{PD}+\frac{FB}{PF}\right)+\left(\frac{CE}{PE}+\frac{DC}{PD}\right)$. Now, in the right-angled triangle BDP, we have $\frac{BD}{PD}=\cot\measuredangle PBD$. Similarly, $\frac{FB}{PF}=\cot\measuredangle PBF$. Now, define $x=90^{\circ}-\measuredangle PBD$; $y=90^{\circ}-\measuredangle PBF$; $x^{\prime}=y^{\prime}=90^{\circ}-\frac{B}{2}$. Then, since $0^{\circ}<\measuredangle PBD<180^{\circ}$, $0^{\circ}<\measuredangle PBF<180^{\circ}$ and $0^{\circ}<\frac{B}{2}<180^{\circ}$, the numbers x, y, x', y' lie in the interval ]-90°; 90°[. Also, $x+y=\left(90^{\circ}-\measuredangle PBD\right)+\left(90^{\circ}-\measuredangle PBF\right)=180^{\circ}-\left(\measuredangle PBD+\measuredangle PBF\right)$ $=180^{\circ}-B=\left(90^{\circ}-\frac{B}{2}\right)+\left(90^{\circ}-\frac{B}{2}\right)=x^{\prime}+y^{\prime}$, and since B < 180° (as B is an angle of triangle ABC), we have x + y = 180° - B > 0°. Thus, $0^{\circ}\leq x+y=x^{\prime}+y^{\prime}$. Finally, $\left|x-y\right|\geq\left|x^{\prime}-y^{\prime}\right|$ (since $x^{\prime}=y^{\prime}$ yields $\left|x^{\prime}-y^{\prime}\right|=0^{\circ}$). Hence, we can apply http://www.mathlinks.ro/Forum/viewtopic.php?t=21384 post #6 Theorem 1 and obtain $\tan x+\tan y\geq\tan x^{\prime}+\tan y^{\prime}$. In other words, $\tan\left(90^{\circ}-\measuredangle PBD\right)+\tan\left(90^{\circ}-\measuredangle PBF\right)\geq\tan\left(90^{\circ}-\frac{B}{2}\right)+\tan\left(90^{\circ}-\frac{B}{2}\right)$. This rewrites as $\cot\measuredangle PBD+\cot\measuredangle PBF\geq\cot\frac{B}{2}+\cot\frac{B}{2}$, what transforms into $\frac{BD}{PD}+\frac{FB}{PF}\geq 2\cot\frac{B}{2}$. Also, as we can easily see, this inequality becomes an equality if and only if x = y, i. e. if and only if 90° - < PBD = 90° - < PBF, i. e. if and only if < PBD = < PBF, i. e. if and only if the point P lies on the angle bisector of the angle ABC. So we have the inequality $\frac{BD}{PD}+\frac{FB}{PF}\geq 2\cot\frac{B}{2}$ with equality if and only if the point P lies on the angle bisector of the angle ABC. Similarly, $\frac{AF}{PF}+\frac{EA}{PE}\geq 2\cot\frac{A}{2}$ with equality if and only if the point P lies on the angle bisector of the angle CAB, and $\frac{CE}{PE}+\frac{DC}{PD}\geq 2\cot\frac{C}{2}$ with equality if and only if the point P lies on the angle bisector of the angle BCA. Thus, we have $\frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF}=\left(\frac{AF}{PF}+\frac{EA}{PE}\right)+\left(\frac{BD}{PD}+\frac{FB}{PF}\right)+\left(\frac{CE}{PE}+\frac{DC}{PD}\right)$ $\geq 2\cot\frac{A}{2}+2\cot\frac{B}{2}+2\cot\frac{C}{2}$, with equality if and only if all the three inequalities $\frac{AF}{PF}+\frac{EA}{PE}\geq 2\cot\frac{A}{2}$, $\frac{BD}{PD}+\frac{FB}{PF}\geq 2\cot\frac{B}{2}$ and $\frac{CE}{PE}+\frac{DC}{PD}\geq 2\cot\frac{C}{2}$ become equalities, i. e. if the point P lies on the angle bisectors of the angles CAB, ABC and BCA, i. e. if the point P is the incenter of triangle ABC. Hence, the value of $\frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF}$ is minimal if the point P is the incenter of triangle ABC. I know this is about 6 times longer than Kalva's / the proposed solution, but it's original, isn't it? Darij

When all points $D, E, F$ are on the segments, we can translate the expression into the sum of six cotangents, and then apply Jensen's inequality (using the fact that $cot(x)$ is convex on $(0,\frac{\pi}{2}]$ ) to three pairs of cotangents to get the lower bound of $2(\cot(A/2)+\cot(B/2)+\cot(C/2))$, with equality when $P$ is the incenter. In attempts to find a way to show the bound for the case when one of $D, E, F$ is beyond the segment I came up with the Cauchy solution as @above, which, apparently, also solves the first case. Anyways, I decided to share my Jensen approach.