Erdős–Mordell inequality

Erdős–Mordell inequality

Let ${\displaystyle P}$ be an arbitrary point P inside a given triangle ${\displaystyle ABC}$, and let ${\displaystyle PL}$, ${\displaystyle PM}$, and ${\displaystyle PN}$ be the perpendiculars from ${\displaystyle P}$ to the sides of the triangles. (If the triangle is obtuse, one of these perpendiculars may cross through a different side of the triangle and end on the line supporting one of the sides.) Then the inequality states that

${\displaystyle PA+PB+PC\geq 2(PL+PM+PN)}$

Let the sides of ABC be a opposite A, b opposite B, and c opposite C; also let PA = p, PB = q, PC = r, dist(P;BC) = x, dist(P;CA) = y, dist(P;AB) = z. First, we prove that

${\displaystyle cr\geq ax+by.}$

This is equivalent to

${\displaystyle {\frac {c(r+z)}{2}}\geq {\frac {ax+by+cz}{2}}.}$

The right side is the area of triangle ABC, but on the left side, r + z is at least the height of the triangle; consequently, the left side cannot be smaller than the right side. Now reflect P on the angle bisector at C. We find that cr ≥ ay + bx for P's reflection. Similarly, bq ≥ az + cx and ap ≥ bz + cy. We solve these inequalities for r, q, and p:

${\displaystyle r\geq (a/c)y+(b/c)x,}$

${\displaystyle q\geq (a/b)z+(c/b)x,}$

${\displaystyle p\geq (b/a)z+(c/a)y.}$

Adding the three up, we get

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

Since the sum of a positive number and its reciprocal is at least 2 by AM–GM inequality, we are finished. Equality holds only for the equilateral triangle, where P is its centroid.

Let ABC be a triangle inscribed into a circle (O) and P be a point inside of ABC. Let D, E, F be the orthogonal projections of P onto BC, CA, AB. M, N, Q be the orthogonal projections of P onto tangents to (O) at A, B, C respectively, then:

${\displaystyle PM+PN+PQ\geq 2(PD+PE+PF)}$

Equality hold if and only if triangle ABC is equilateral (Dao, Nguyen & Pham 2016; Marinescu & Monea 2017)

Let ${\displaystyle A_{1}A_{2}...A_{n}}$ be a convex polygon, and ${\displaystyle P}$ be an interior point of ${\displaystyle A_{1}A_{2}...A_{n}}$. Let ${\displaystyle R_{i}}$ be the distance from ${\displaystyle P}$ to the vertex ${\displaystyle A_{i}}$ , ${\displaystyle r_{i}}$ the distance from ${\displaystyle P}$ to the side ${\displaystyle A_{i}A_{i+1}}$, ${\displaystyle w_{i}}$ the segment of the bisector of the angle ${\displaystyle A_{i}PA_{i+1}}$ from ${\displaystyle P}$ to its intersection with the side ${\displaystyle A_{i}A_{i+1}}$ then (Lenhard 1961):

${\displaystyle \sum _{i=1}^{n}R_{i}\geq \left(\sec {\frac {\pi }{n}}\right)\sum _{i=1}^{n}w_{i}\geq \left(\sec {\frac {\pi }{n}}\right)\sum _{i=1}^{n}r_{i}}$

• List of triangle inequalities

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• Alexander Bogomolny, "Erdös-Mordell Inequality", from Cut-the-Knot.