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Romania Team Selection Test
1992 Romania Team Selection Test
4
x_1^2 +x_2^2+...+x_n^2= 1, [x_1 +x_2 +...+x_n] = m, x_1 +x_2 +...+x_m \ge 1
x_1^2 +x_2^2+...+x_n^2= 1, [x_1 +x_2 +...+x_n] = m, x_1 +x_2 +...+x_m \ge 1
Source: Romania BMO TST 1992 p4
February 19, 2020
Sum
inequalities
algebra
Problem Statement
Let
x
1
,
x
2
,
.
.
.
,
x
n
x_1,x_2,...,x_n
x
1
,
x
2
,
...
,
x
n
be real numbers with
1
≥
x
1
≥
x
2
≥
.
.
.
≥
x
n
≥
0
1 \ge x_1 \ge x_2\ge ... \ge x_n \ge 0
1
≥
x
1
≥
x
2
≥
...
≥
x
n
≥
0
and
x
1
2
+
x
2
2
+
.
.
.
+
x
n
2
=
1
x_1^2 +x_2^2+...+x_n^2= 1
x
1
2
+
x
2
2
+
...
+
x
n
2
=
1
. If
[
x
1
+
x
2
+
.
.
.
+
x
n
]
=
m
[x_1 +x_2 +...+x_n] = m
[
x
1
+
x
2
+
...
+
x
n
]
=
m
, prove that
x
1
+
x
2
+
.
.
.
+
x
m
≥
1
x_1 +x_2 +...+x_m \ge 1
x
1
+
x
2
+
...
+
x
m
≥
1
.
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