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Indonesia TST
2010 Indonesia TST
4
n! + 1 \ge \sum_{j \in J}a_j >\sqrt {n! + (n - 1)n}
n! + 1 \ge \sum_{j \in J}a_j >\sqrt {n! + (n - 1)n}
Source: 2010 Indonesia TST stage 2 test 4 p4
December 16, 2020
inequalities
algebra
Problem Statement
Given a positive integer
n
n
n
and
I
=
{
1
,
2
,
.
.
.
,
k
}
I = \{1, 2,..., k\}
I
=
{
1
,
2
,
...
,
k
}
with
k
k
k
is a positive integer. Given positive integers
a
1
,
a
2
,
.
.
.
,
a
k
a_1, a_2, ..., a_k
a
1
,
a
2
,
...
,
a
k
such that for all
i
∈
I
i \in I
i
∈
I
:
1
≤
a
i
≤
n
1 \le a_i \le n
1
≤
a
i
≤
n
and
∑
i
=
1
k
a
i
≥
2
(
n
!
)
.
\sum_{i=1}^k a_i \ge 2(n!).
i
=
1
∑
k
a
i
≥
2
(
n
!)
.
Show that there exists
J
⊆
I
J \subseteq I
J
⊆
I
such that
n
!
+
1
≥
∑
j
∈
J
a
j
>
n
!
+
(
n
−
1
)
n
n! + 1 \ge \sum_{j \in J}a_j >\sqrt {n! + (n - 1)n}
n
!
+
1
≥
j
∈
J
∑
a
j
>
n
!
+
(
n
−
1
)
n
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