Erdös-Mordell Inequality

In 1935, the following problem proposal appeared in the "Advanced Problems" section of the American Mathematical Monthly:

3740. Proposed by Paul Erdös, The University of Manchester, England.

From a point O inside a given triangle ABC the perpendiculars OP, OQ, OR are drawn to its sides. Prove that

OA + OB + OC ≥ 2(OP + OQ + OR).

A trigonometric solution by L. J. Mordell and D. F. Barrow was published in 1937. The inequality became known as the Erdös-Mordell Inequality or Erdös-Mordell Theorem.

Numerous additional proofs have been published since. The most elementary yet is of the recent vintage and is due to Claudi Alsina and Roger B. Nelsen.

In the notations of the above diagram the inequality appears as

x + y + z ≥ 2(p + q + r).

The proof of the inequality is based on the following

Lemma

For the quantities x, y, z, p, q, r in ΔABC, we have ax ≥ br + cq, by ≥ ar + cp, and cz ≥ aq + bp.

Proof of Lemma

For the proof we construct a trapezoid as shown. The diagram makes the first inequality ax ≥ br + cq obvious. The other two are shown similarly.

(That we do have a trapezoid follows from counting the angles at vertex A: they do sum up to 180°.)

The Erdös-Mordell Inequality

If O is a point within a triangle ABC whose distances to the vertices are x, y, and z, then

x + y + z ≥ 2(p + q + r).

Proof

From the lemma we have ax ≥ br + cq, by ≥ ar + cp, and cz ≥ aq + bp. Adding these three inequalities yields

x + y + z ≥ (b/a + a/b)r + (c/a + a/c)q + (c/b + b/c)p.

But the arithmetic mean-geometric mean inequality insures that the coefficients of p, q, and r are each at least 2, from which the desired result follows.

Observe that the three inequalities in the lemma are equalities if and only if O is the circumcenter of ΔABC, for in this case the trapezoids become rectangles.

Remark

If the arithmetic mean-geometric mean inequality is applied directly to the identities in Lemma, one obtains another relationship:

xyz ≥ 8pqr.

Note

In a 1957 paper (D. K. Kazarinoff's inequality for tetrahedra, Michigan Math. J., 4 (1957), pp 99-104) Nicholas D. Kazarinoff proved an analogue of the Erdös-Mordell inequality for tetrahedra: Let S be a tetrahedron, and P a point not exterior to S. Let the distances from P to the vertices and to the faces be denoted by R_i and r_i respectively. The following analogue of the Erdös-Mordell inequality holds for any tetrahedron whose circumcenter is not an exterior point

ΣR_i / Σr_i > 2√2.

and 2√2 is the greatest lower bound.

Reference

Claudi Alsina and Roger B. Nelsen, A Visual Proof of the Erdös-Mordell Inequality, Forum Geometricorum, Volume 7 (2007) 99-102
C. Alsina, R. B. Nelsen, Charming Proofs, MAA, 2010, pp. 82-84