MathLabs

Problem 1

Let PP be a point inside a given triangle ABCABC, and let DD, EE, FF be the feet of the perpendiculars from PP to the lines BCBC, CACA, ABAB respectively. Find all positions of PP for which BCPD+CAPE+ABPF\dfrac{BC}{PD} + \dfrac{CA}{PE} + \dfrac{AB}{PF} is least.
Step 5 of 5: Identify the equality case
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PD=PE=PF  ⟺  P=I (the incenter)PD = PE = PF \iff P = I \text{ (the incenter)}
Detailed analysis

Equality in Cauchy–Schwarz holds exactly when PD=PE=PFPD=PE=PF, i.e. when PP is equidistant from the three sides. The points equidistant from all three sidelines are the incenter and the three excenters; only the incenter lies inside the triangle, so the minimum is attained uniquely at P=IP=I.

Common mistake. The three excenters also satisfy PD=PE=PFPD=PE=PF in the sense of distances to the side lines, but they lie outside △ABC\triangle ABC, so the constraint "PP inside the triangle" is what singles out the incenter.