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National and Regional Contests
China Contests
China Girls Math Olympiad
2016 China Girls Math Olympiad
4
4
Part of
2016 China Girls Math Olympiad
Problems
(1)
Chinese Girls Mathematical Olympiad 2016, Problem 4
Source: China Beijing ,12 Aug 2016
8/12/2016
Let
n
n
n
is a positive integers ,
a
1
,
a
2
,
⋯
,
a
n
∈
{
0
,
1
,
⋯
,
n
}
a_1,a_2,\cdots,a_n\in\{0,1,\cdots,n\}
a
1
,
a
2
,
⋯
,
a
n
∈
{
0
,
1
,
⋯
,
n
}
. For the integer
j
j
j
(
1
≤
j
≤
n
)
(1\le j\le n)
(
1
≤
j
≤
n
)
,define
b
j
b_j
b
j
is the number of elements in the set
{
i
∣
i
∈
{
1
,
⋯
,
n
}
,
a
i
≥
j
}
\{i|i\in\{1,\cdots,n\},a_i\ge j\}
{
i
∣
i
∈
{
1
,
⋯
,
n
}
,
a
i
≥
j
}
.For example ：When
n
=
3
n=3
n
=
3
,if
a
1
=
1
,
a
2
=
2
,
a
3
=
1
a_1=1,a_2=2,a_3=1
a
1
=
1
,
a
2
=
2
,
a
3
=
1
,then
b
1
=
3
,
b
2
=
1
,
b
3
=
0
b_1=3,b_2=1,b_3=0
b
1
=
3
,
b
2
=
1
,
b
3
=
0
.
(
1
)
(1)
(
1
)
Prove that
∑
i
=
1
n
(
i
+
a
i
)
2
≥
∑
i
=
1
n
(
i
+
b
i
)
2
.
\sum_{i=1}^{n}(i+a_i)^2\ge \sum_{i=1}^{n}(i+b_i)^2.
i
=
1
∑
n
(
i
+
a
i
)
2
≥
i
=
1
∑
n
(
i
+
b
i
)
2
.
(
2
)
(2)
(
2
)
Prove that
∑
i
=
1
n
(
i
+
a
i
)
k
≥
∑
i
=
1
n
(
i
+
b
i
)
k
,
\sum_{i=1}^{n}(i+a_i)^k\ge \sum_{i=1}^{n}(i+b_i)^k,
i
=
1
∑
n
(
i
+
a
i
)
k
≥
i
=
1
∑
n
(
i
+
b
i
)
k
,
for the integer
k
≥
3.
k\ge 3.
k
≥
3.
algebra
inequalities