IMO 1981 SL 15

Origin: GBR

Problem

Find the point P inside the triangle ABC for which BC PD + CA PE + AB PF is minimal, where PD, PE, PF are the perpendiculars from P to BC, CA, AB respectively.

Solution

Set BC = a, CA = b, AB = c, and denote the area of \triangleABC by P, and a/PD + b/PE + c/PF by S. Since a \cdot PD + b \cdot PE + c \cdot PF = 2P, by the Cauchy–Schwarz inequality we have 2PS = (a \cdot PD + b \cdot PE + c \cdot PF)  a PD + b PE + c PF  \geq(a + b + c)2, with equality if and only if PD = PE = PF, i.e., P is the incenter of \triangleABC. In that case, S attains its minimum:

Smin = (a + b + c)2 2P .