Advanced book on Mathematics Olympiad

Geometry and Trigonometry 657

So we have to prove that

∏n

k= 0

1 +bk

1 −bk

≥nn+^1.

The inequality from the statement implies

1 +bk≥

∑

l   =k

( 1 −bl), k= 0 , 1 ,...,n.

Also, the conditionak∈( 0 ,π 2 )implies− 1 <bk<1,k= 0 , 1 ,...,n, so the numbers
1 −bkare all positive. To obtain their product, it is natural to apply the AM–GM inequality
to the right-hand side of the above inequality, and obtain

1 +bk≥nn

√∏

l   =k

( 1 −bl), k= 0 , 1 ,...,n.

Multiplying all these inequalities yields

∏n

k= 0

( 1 +bk)≥nn+^1 n

√√

√√∏n

l= 0

( 1 −bl)n.

Hence

∏n

k= 0

1 +bk

1 −bk

≥nn+^1 ,

as desired.

(USA Mathematical Olympiad, 1998, proposed by T. Andreescu)

666.If we multiply the denominator and the numerator of the left-hand side by cost, and

of the right-hand side by cosnt, we obtain the obvious equality

(

eit

e−it

)n

=

eint

e−int

.

667.Using the de Moivre formula, we obtain

( 1 +i)n=

[√

2

(

cos

π

4

+isin

π

4

)]n

= 2 n/^2

(

cos

nπ

4

+isin

nπ

4

)

.

Expanding( 1 +i)nand equating the real parts on both sides, we deduce the identity from

the statement.