Coxeter Exercise 4 of Section 1.3 in "Introduction To Geometry"

In this note we do the exercise listed in the title according to the hint given by the author.

   In the last blog entry it was shown that right triangles BOR and RPP1 in the diagram above were similar by constructing a circle whose center was the midpoint of segment OB, which included points B, P, O, R, and thus had inscribed angles BPR and BOR both subtended by chord BR and therefore equal to each other. We will call this circle [BPOR]. This result is the key to the exercise which is to prove the Erdos-Mordell Theorem: If O is any point inside a triangle ABC and P, Q, R are the feet of the perpendiculars from O upon the respective sides BC, CA, AB, then OA + OB + OC ≥ 2(OP + OQ + OR).
   The hint supplied by Coxeter is this: Let P1 and P2 be the feet of perpendiculars from R and Q upon BC. Define analogous points Q1 and Q2, R1 and R2 on the other sides. Using the similarity of the triangles PRP1 and OBR, express P1P in terms of RP, OR, and OB. After substituting such expressions into OA + OB + OC ≥ OA(P1P + PP2)/RQ + OB(Q1Q + QQ2)/PR + OC(R1R + RR2)/QP, collect the terms involving OP, OQ, OR, respectively.
   In the diagram above the points P1 and P2 are shown along with the feet of the perpendiculars from the arbitrary point O in ABC. Including the perpendiculars from P to Q1, R to Q2, P to R1, and Q to R2 makes for a pretty complicated drawing. It is suggested the reader draw the other two cases separately in order to better picture the argument below. Note: To play around with the parameters of the diagram, try different (sliding) values of the variables defined at the left in this diagram.
   Just as we constructed circle [BPOR] and established the similarity of triangles BOR and RPP1, we can construct five other such circles and pairs of similar triangles. Thus, circle [CPOQ] yields the similarity of triangles COQ and QPP2, circle [CQOP] yields COP similar to PQQ1, circle [AQOR] yields AOR similar to RQQ2, circle [BROP] yields BOP similar to PRR1, and circle [AROQ] yields AOQ similar to QRR2.
   From these similarities we can write:
PP1/OR = RP1/BR = RP/OB, so PP1 = (OR x RP)/OB
QP2/CG = PP2/OQ=PQ/OC, so PP2 = (OQ x PQ)/OC
PQ1/PC = QQ1/OP = PQ/OC, so QQ1 = (OP x PQ)/OC
RQ2/RA = QQ2/OR = RQ/OA, so QQ2 = (OR x RQ)/OA
PR1/BP = RR1/OP = RP/OB, so RR1 = (OP x RP)/OB
QR2/AQ = RR2/OQ = RQ/OA, so RR2 = (OQ x RQ)/OA.
   Also notice that in the quadrilateral RP1P2Q, the distance between parallel segments RP1 and QP2 is P1P2 and therefore RQ ≥ P1P2 = P1P + PP2, or (P1P + PP2)/RQ ≤ 1. Analogously, we get
(Q1Q + QQ2)/PR ≤ 1 and (R1R + RR2)/QP ≤ 1. Thus,
OA + OB + OC ≥ OA (P1P + PP2)/RQ + OB (Q1Q + QQ2)/PR + OC (R1R + RR2)/QP.
   Now substitute the expressions obtained by the similarities into the right hand side of this last expression and collect terms to get
OA + OB + OC ≥ OR[OA.RP/RQ.OB + OB.RQ/RP.OA] + OQ[OA.PQ/RQ.OC + OC.RQ/PQ.OA] + OP[OB.PQ/RP.OC + OC.RP/PQ.OB] which is of the form
OA + OB + OC ≥ OR[z + 1/z] + OQ[y + 1/y] + OP[x + 1/x], where x, y, z > 0.
   If f(u) = u + 1/u, u > 0, then f'(u) = 1 - 1/u^2 = 0 when u=1, f''(1) > 0, so the minimum value of f is f(1) = 2. Thus OA + OB + OC ≥ 2(OR + OQ + OP).