Let ABC be an acute triangle such that AB<AC and AB<BC. Let P be a point on the segment BC such that ∠APB=∠BAC. The tangent to the circumcircle of triangle ABC at A meets the circumcircle of triangle APB at Q=A. Let Q′ be the reflection of Q with respect to the midpoint of AB. The line PQ meets the segment AQ′ at S. Prove that
AB1+AC1>CS1.Proposed by Nikola Velov geometric inequalityinequalitiesgeometrygeometric transformationreflection