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JOM 2023
4
Minimum value of sum
Minimum value of sum
Source: JOM 2023 P4
February 20, 2023
algebra
Problem Statement
Given
n
n
n
positive real numbers
x
1
,
x
2
,
x
3
,
.
.
.
,
x
n
x_1,x_2,x_3,...,x_n
x
1
,
x
2
,
x
3
,
...
,
x
n
such that
(
1
+
1
x
1
)
(
1
+
1
x
2
)
.
.
.
(
1
+
1
x
n
)
=
(
n
+
1
)
n
\left (1+\frac{1}{x_1}\right )\left(1+\frac{1}{x_2}\right)...\left(1+\frac{1}{x_n}\right)=(n+1)^n
(
1
+
x
1
1
)
(
1
+
x
2
1
)
...
(
1
+
x
n
1
)
=
(
n
+
1
)
n
Determine the minimum value of
x
1
+
x
2
+
x
3
+
.
.
.
+
x
n
x_1+x_2+x_3+...+x_n
x
1
+
x
2
+
x
3
+
...
+
x
n
.Proposed by Loh Kwong Weng
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