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max (AP, BP, CP)>=\sqrt{d_a^2+d_b^2+d_c^2} where d_i distances of interior P

Source: Balkan MO Shortlist 2008 G8

April 6, 2020
geometric inequalitygeometrymaxdistance

Problem Statement

Let PP be a point in the interior of a triangle ABCABC and let da,db,dcd_a,d_b,d_c be its distances to BC,CA,ABBC,CA,AB respectively. Prove that max (AP,BP,CP)≥da2+db2+dc2(AP, BP, CP) \ge \sqrt{d_a^2+d_b^2+d_c^2}
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