66th Belarusian Mathematical Olympiad

Question

Let a, b, c, d, x, y denote the lengths of the sides AB, BC, CD, DA and the diagonals AC, BD of a cyclic quadrilateral ABCD, respectively. (1a + 1c)^2 + (1b + 1d)^2 8 (1x^2 + 1y^2). FIGURE_0

Accepted Answer

(Solution by Y. Dubovik, U. Kazlouski.) By the Cosine Law for the triangles

D A B

and

D C B

,

y^{2} = a^{2} + d^{2} - 2 a d cos A, (1)

y^{2} = b^{2} + c^{2} + 2 b c cos A . (2)

Multiplying (1) and (2) by

b c

and

a d

, respectively, and summing the obtained equalities, we get

(b c + a d) y^{2} = (a^{2} + d^{2}) b c + (b^{2} + c^{2}) a d

\geq 2 a d b c + 2 b c a d = 4 ab c d .

\frac{8}{y ^{2}} \leq \frac{2 ( b c + a d )}{ab c d} = \frac{2}{a d} + \frac{2}{b c} . (3)

In the same way we can obtain

\frac{8}{x ^{2}} \leq \frac{2}{ab} + \frac{2}{c d} . (4)

So,

8 (\frac{1}{x ^{2}} + \frac{1}{y ^{2}}) \leq \frac{2}{a d} + \frac{2}{b c} + \frac{2}{ab} + \frac{2}{c d} = 2 (\frac{1}{a} + \frac{1}{c}) (\frac{1}{b} + \frac{1}{d}) \leq (\frac{1}{a} + \frac{1}{c})^{2} + (\frac{1}{b} + \frac{1}{d})^{2},

as required.