Advanced book on Mathematics Olympiad

372 Algebra

This is equivalent to

0 <(a+b+c)(−a+b+c)(a−b+c)(a−b−c).

It follows that−a+b+c,a−b+c,a−b−care all positive, becausea+b+c>0,
and no two of the factors could be negative, for in that case the sum of the three numbers
would also be negative. Done.

117.The first idea is to simplify the problem and prove separately the inequalities|AB−
CD|≥|AC−BD|and|AD−BC|≥|AC−BD|. Because of symmetry it suffices to
prove the first.
LetMbe the intersection of the diagonalsACandBD. For simplicity, letAM=x,
BM=y,AB=z. By the similarity of trianglesMABandMDCthere exists a positive
numberksuch thatDM=kx,CM =ky,andCD=kz(Figure 60). Then

|AB−CD|=|k− 1 |z

and

|AC−BD|=|(kx+y)−(ky+x)|=|k− 1 |·|x−y|.

By the triangle inequality,|x−y|≤z, which implies|AB−CD|≥|AC−BD|,
completing the proof.

A

BC

D

z

x

y

M

kx

ky

kz

Figure 60

(USA Mathematical Olympiad, 1999, proposed by T. Andreescu, solution by
P.R. Loh)

118.We induct onm. Whenm=1 there is nothing to prove. Now assume that the
inequality holds form−1 isometries and let us prove that it holds formisometries.
DefineV=

∏m− 1

i= 1 ViandW=

∏m− 1
i= 1 Wi. BothVandWare isometries. For a vectorx
with‖x‖≤1,