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Poland Contests
Poland - Second Round
1981 Poland - Second Round
1
\sqrt{\sum (x_j^2+y_j^2)} <=1 / \sqrt{2} \sum \sqrt{x_j^2+y_j^2}.
\sqrt{\sum (x_j^2+y_j^2)} <=1 / \sqrt{2} \sum \sqrt{x_j^2+y_j^2}.
Source: Polish MO Recond Round 1981 p1
September 9, 2024
algebra
inequalities
Problem Statement
Prove that for any real numbers
x
1
,
x
2
,
…
,
x
1981
x_1, x_2, \ldots, x_{1981}
x
1
,
x
2
,
…
,
x
1981
,
y
1
,
y
2
,
…
,
y
1981
y_1, y_2, \ldots, y_{1981}
y
1
,
y
2
,
…
,
y
1981
such that
∑
j
=
1
1981
x
j
=
0
\sum_{j=1}^{1981} x_j = 0
∑
j
=
1
1981
x
j
=
0
,
∑
j
=
1
1981
y
j
=
0
\sum_{j=1}^{1981} y_j = 0
∑
j
=
1
1981
y
j
=
0
the inequality occurs
∑
j
=
1
1981
(
x
j
2
+
y
j
2
)
≤
1
2
∑
j
=
1
1981
x
j
2
+
y
j
2
.
\sqrt{\sum_{j=1}^{1981} (x_j^2+y_j^2)} \leq \frac{1}{\sqrt{2}} \sum_{j=1}^{1981} \sqrt{x_j^2+y_j^2}.
j
=
1
∑
1981
(
x
j
2
+
y
j
2
)
≤
2
1
j
=
1
∑
1981
x
j
2
+
y
j
2
.
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