Erdős Problem 898: the Erdős--Mordell inequality
Statement.
Let $P$ lie strictly inside a nondegenerate triangle $ABC$. Write $R_A=PA,R_B=PB,R_C=PC$, and let $r_A,r_B,r_C$ be the perpendicular distances from $P$ to the supporting lines $BC,CA,AB$, respectively. The assertion is \[
R_A+R_B+R_C\geq2(r_A+r_B+r_C).
\] Distances to the side lines are intended even when the foot lies on an extension of a side.
A local weighted inequality.
Put $a=BC,b=CA,c=AB$, let $\alpha=\angle BAC$, and let $\theta=\angle BAP\in(0,\alpha)$. Then \[
r_C=R_A\sin\theta,\qquad
r_B=R_A\sin(\alpha-\theta).
\] Consequently \[
\frac{c r_B+b r_C}{R_A}
=c\sin\alpha\cos\theta+(b-c\cos\alpha)\sin\theta
\leq\sqrt{c^2\sin^2\alpha+(b-c\cos\alpha)^2}
=a.
\] The inequality is Cauchy--Schwarz, and the last identity is the cosine rule. Thus \[
R_A\geq(c/a)r_B+(b/a)r_C.
\tag{1}
\] No acute-angle assumption is involved.
Summing cyclically.
Apply $1$ at each vertex and collect the three side distances: \[
\begin{split}
R_A+R_B+R_C
&\geq (c/b+b/c)r_A
+(c/a+a/c)r_B
+(b/a+a/b)r_C\\
&\geq2(r_A+r_B+r_C),
\end{split}
\] because $u/u'+u'/u\geq2$ for positive $u,u'$. This proves the full inequality.
Equality and sharpness.
Since all $r_A,r_B,r_C$ are positive, equality in the last line forces $a=b=c$. For an equilateral triangle the Cauchy--Schwarz equality at $A$ forces $AP$ to be its internal angle bisector, and the analogous conditions at the other vertices force $P$ to be the centre. Conversely that centre has $R_A=R_B=R_C=2r_A=2r_B=2r_C$, giving equality. Thus the constant two cannot be increased.
Attribution and assessment.
This is a standard elementary proof of the classical Erdős--Mordell theorem, not a new result. The current statement and historical attribution are at
https://www.erdosproblems.com/898, checked 24 September 2026. The proof is complete using only the cosine rule and Cauchy--Schwarz; no external resolution theorem or local Lean verification is being assumed.
Completion estimate.
COMPLETION: 100%