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Contests
National and Regional Contests
Moldova Contests
JBMO TST - Moldova
2004 Junior Balkan Team Selection Tests - Moldova
2
sum \sqrt{a_i^2+1}>= \sqrt{2n(a_1 + a_2 +...+ a_n}}
sum \sqrt{a_i^2+1}>= \sqrt{2n(a_1 + a_2 +...+ a_n}}
Source: 2004 Moldova JBMO TST p2
February 20, 2021
algebra
inequalities
Problem Statement
Let
n
∈
N
∗
n \in N^*
n
∈
N
∗
. Let
a
1
,
a
2
.
.
.
,
a
n
a_1, a_2..., a_n
a
1
,
a
2
...
,
a
n
be real such that
a
1
+
a
2
+
.
.
.
+
a
n
≥
0
a_1 + a_2 +...+ a_n \ge 0
a
1
+
a
2
+
...
+
a
n
≥
0
. Prove the inequality
a
1
2
+
1
+
a
2
2
+
1
+
.
.
.
+
a
1
2
+
1
≥
2
n
(
a
1
+
a
2
+
.
.
.
+
a
n
)
\sqrt{a_1^2+1}+\sqrt{a_2^2+1}+...+\sqrt{a_1^2+1}\ge \sqrt{2n(a_1 + a_2 +...+ a_n )}
a
1
2
+
1
+
a
2
2
+
1
+
...
+
a
1
2
+
1
≥
2
n
(
a
1
+
a
2
+
...
+
a
n
)
.
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