Erdős–Mordell inequality

In Euclidean geometry, the Erdős–Mordell inequality states that for any triangle ABC and point P inside ABC, the sum of the distances from P to the sides is less than or equal to half of the sum of the distances from P to the vertices. It is named after Paul Erdős and Louis Mordell. posed the problem of proving the inequality; a proof was provided two years later by. This solution was however not very elementary. Subsequent simpler proofs were then found by, , and.

Barrow's inequality is a strengthened version of the Erdős–Mordell inequality in which the distances from P to the sides are replaced by the distances from P to the points where the angle bisectors of ∠APB, ∠BPC, and ∠CPA cross the sides. Although the replaced distances are longer, their sum is still less than or equal to half the sum of the distances to the vertices.

Statement
Let $$P$$ be an arbitrary point P inside a given triangle $$ABC$$, and let $$PL$$, $$PM$$, and $$PN$$ be the perpendiculars from $$P$$ 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
 * $$PA+PB+PC\geq 2(PL+PM+PN)$$

Proof
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


 * $$cr\geq ax+by.$$

This is equivalent to


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

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:


 * $$r\geq (a/c)y+(b/c)x,$$


 * $$q\geq (a/b)z+(c/b)x,$$


 * $$p\geq (b/a)z+(c/a)y.$$

Adding the three up, we get



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.

Another strengthened version
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:
 * $$ PM+PN+PQ \ge 2(PD+PE+PF)$$

Equality hold if and only if triangle ABC is equilateral

A generalization
Let $$A_1A_2...A_n$$ be a convex polygon, and $$P$$ be an interior point of $$A_1A_2...A_n$$. Let $$R_i$$ be the distance from $$P$$ to the vertex $$A_i$$, $$r_i$$ the distance from $$P$$ to the side $$A_iA_{i+1}$$, $$w_i$$ the segment of the bisector of the angle $$A_iPA_{i+1}$$ from $$P$$ to its intersection with the side $$A_iA_{i+1}$$ then :
 * $$ \sum_{i=1}^{n}R_i \ge \left(\sec{\frac{\pi}{n}}\right)\sum_{i=1}^{n} w_i \ge \left(\sec{\frac{\pi}{n}}\right)\sum_{i=1}^{n} r_i $$

In absolute geometry
In absolute geometry the Erdős–Mordell inequality is equivalent, as proved in, to the statement that the sum of the angles of a triangle is less than or equal to two right angles.