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\sqrt{S_1} + \sqrt{S_2} <= \sqrt{S} areas in quadrilateral

Source: 1986 Swedish Mathematical Competition p2

March 28, 2021

geometryGeometric Inequalitiestriangle area

Problem Statement

The diagonals

AC

and

BD

of a quadrilateral

ABCD

intersect at

O

. If

S_1

and

S_2

are the areas of triangles

AOB

and

COD

and S that of

ABCD

, show that

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

. Prove that equality holds if and only if

AB

and

CD

are parallel.