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\sqrt{S_1}+\sqrt{S_2}\le \sqrt{S} in convex ABCD

Source: 2006 Spanish Mathematical Olympiad P6

July 20, 2018

geometrygeometric inequalityconvex quadrilateraldiagonalssquare rootsareas

Problem Statement

The diagonals

AC

and

BD

of a convex quadrilateral

ABCD

intersect at

E

. Denotes by

S_1,S_2

and

S

the areas of the triangles

ABE

CDE

and the quadrilateral

ABCD

respectively. Prove that

\sqrt{S_1}+\sqrt{S_2}\le \sqrt{S}

. When equality is reached?