# 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

$OA + OB + OC \ge 2(OP + OQ + OR).$

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

$x + y + z \ge 2(p + q + r).$

The proof of the inequality is based on the following

### Lemma

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

### Proof of Lemma For the proof we construct a trapezoid as shown. The diagram makes the first inequality $ax \ge br + cq\,$ 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 $180^{\circ}.)$ ### The Erdös-Mordell Inequality

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

$x + y + z \ge 2(p + q + r).$

### Proof

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

$\displaystyle x + y + z \ge \left(\frac{b}{a} + \frac{a}{b}\right)r + \left(\frac{c}{a} + \frac{a}{c}\right)q + \left(\frac{c}{b} + \frac{b}{c}\right)p.$

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 $\Delta ABC,\,$ 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:

$xyz \ge 8pqr.$

### Note 1

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

$\displaystyle\frac{\displaystyle\sum_{cycl}R_i }{\displaystyle\sum_{cycl}r_i} \ge 2\sqrt{2}$.

and $2\sqrt{2}\,$ is the greatest lower bound.

### Note 2

Another extension of the Erdös-Mordell inequality dates back to 1961 (L. Fejes conjectured, H. C. Lenhard proved) that for $n-\text{gon}:\,$

Let $A_1A_2\ldots A_nA_{n+1}=A_1\,$ be a convex polygon and $P\,$ a point in its interior. Let $R_k=PA_k\,$ while $r_k\,$ denote the distance from $P\,$ to the line $A_kA_{k=1}.\,$ Then

$\displaystyle\cos\frac{\pi}{n}\sum_{k=1}^nR_k\ge\sum_{k=1}^nr_k.$

Marian Dinca gave a short proof in 1998 and then 1999. More recently (2009) Marian Dinca devised an inequality for the polygons that are not necessarily flat as well as a more basic inequality:

For every $x_k\ge 0,\,$ $1\le k\le n,\,$ $x_{n+1}=x_1\,$ and $\alpha_k\in (0,\pi),\,$ with $\displaystyle\sum_{k=1}^n\alpha_k=\pi,\,$ the following inequality holds

$\displaystyle\cos\frac{\pi}{n}\sum_{k=1}^nx^2_k\ge\sum_{k=1}^nx_kx_{k+1}\cos\alpha_k.$

The Erdös-Mordell inequality follows from that but there is a more general result:

Suppose $\mathbf{x}=(x_1,\ldots,x_n)\,$ and $\mathbf{y}=(y_1,\dots,y_n)\,$ are vectors in $\mathbb{R}^n,\,$ endowed with the inner product

$\displaystyle\langle\mathbf{x},\mathbf{y}\rangle=||\mathbf{x}||\cdot ||\mathbf{y}||\cos\theta=\sum_{k=1}^nx_ky_k,$

where $\theta\,$ is the angle between the vectors $\mathbf{x}\,$ and $\mathbf{y},\,$ while

$||\mathbf{x}||=\sqrt{x^2_1+x^2_2+\ldots+x^2_n}.$

Then the following inequility of the Erdös-Mordell type holds:

$\displaystyle\cos\frac{\pi}{n}\sum_{k=1}^n||\mathbf{X_k}||\ge\sum_{k=1}^n\langle\mathbf{X_k},\mathbf{X_{k+1}}\rangle=\sum_{k=1}^n||\mathbf{X_k}||\cdot ||\mathbf{X_{k+1}}||\cos\theta_k,$

$\theta_k\,$ is the angle between $\mathbf{X_k}\,$ and $\mathbf{X_{k+1}}\,$ and $\displaystyle\sum_{k=1}^n\theta_k=\pi.$

### Reference

1. Claudi Alsina and Roger B. Nelsen, A Visual Proof of the Erdös-Mordell Inequality, Forum Geometricorum, Volume 7 (2007) 99-102
2. C. Alsina, R. B. Nelsen, Charming Proofs, MAA, 2010, pp. 82-84
3. , M. Dinca, A SIMPLE PROOF OF THE ERDOS-MORDELL INEQUALITY FOR POLYGONS IN N-DIMENSIONAL SPACES, Articole si Note Matematice, 2009  