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2017 Singapore MO Open
2
sum (a_i-b_i) (a_i (sum a_1^2)^{p/2}-b_i (sum b_i^2)^{p/2} >=0
sum (a_i-b_i) (a_i (sum a_1^2)^{p/2}-b_i (sum b_i^2)^{p/2} >=0
Source: Singapore Open Math Olympiad 2017 2nd Round p2 SMO
March 26, 2020
inequalities
algebra
Problem Statement
Let
a
1
,
a
2
,
.
.
.
,
a
n
,
b
1
,
b
2
,
.
.
.
,
b
n
,
p
a_1,a_2,...,a_n,b_1,b_2,...,b_n,p
a
1
,
a
2
,
...
,
a
n
,
b
1
,
b
2
,
...
,
b
n
,
p
be real numbers with
p
>
−
1
p >- 1
p
>
−
1
. Prove that
∑
i
=
1
n
(
a
i
−
b
i
)
(
a
i
(
a
1
2
+
a
2
2
+
.
.
.
+
a
n
2
)
p
/
2
−
b
i
(
b
1
2
+
b
2
2
+
.
.
.
+
b
n
2
)
p
/
2
)
≥
0
\sum_{i=1}^{n}(a_i-b_i)\left(a_i (a_1^2+a_2^2+...+a_n^2)^{p/2}-b_i (b_1^2+b_2^2+...+b_n^2)^{p/2}\right)\ge 0
i
=
1
∑
n
(
a
i
−
b
i
)
(
a
i
(
a
1
2
+
a
2
2
+
...
+
a
n
2
)
p
/2
−
b
i
(
b
1
2
+
b
2
2
+
...
+
b
n
2
)
p
/2
)
≥
0
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