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Baltic Way
2003 Baltic Way
5
Sequence Inequality [Baltic Way 2003]
Sequence Inequality [Baltic Way 2003]
Source:
November 6, 2010
inequalities
induction
algebra proposed
algebra
Problem Statement
The sequence
(
a
n
)
(a_n)
(
a
n
)
is defined by
a
1
=
2
a_1=\sqrt{2}
a
1
=
2
,
a
2
=
2
a_2=2
a
2
=
2
, and
a
n
+
1
=
a
n
a
n
−
1
2
a_{n+1}=a_na_{n-1}^2
a
n
+
1
=
a
n
a
n
−
1
2
for
n
≥
2
n\ge 2
n
≥
2
. Prove that for every
n
≥
1
n\ge 1
n
≥
1
(
1
+
a
1
)
(
1
+
a
2
)
⋯
(
1
+
a
n
)
<
(
2
+
2
)
a
1
a
2
⋯
a
n
.
(1+a_1)(1+a_2)\cdots (1+a_n)<(2+\sqrt{2})a_1a_2\cdots a_n.
(
1
+
a
1
)
(
1
+
a
2
)
⋯
(
1
+
a
n
)
<
(
2
+
2
)
a
1
a
2
⋯
a
n
.
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