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Kvant 2024
M2797
M2797
Part of
Kvant 2024
Problems
(1)
Kvant M2797 inequality
Source: Kvant Magazine No. 5-6 2024 M2797
8/25/2024
For real numbers
0
≤
a
1
≤
a
2
≤
.
.
.
≤
a
n
0 \leq a_1 \leq a_2 \leq ... \leq a_n
0
≤
a
1
≤
a
2
≤
...
≤
a
n
and
0
≤
b
1
≤
b
2
≤
.
.
.
≤
b
n
0 \leq b_1 \leq b_2 \leq ... \leq b_n
0
≤
b
1
≤
b
2
≤
...
≤
b
n
prove that
(
a
1
1
⋅
2
+
a
2
2
⋅
3
+
.
.
.
+
a
n
n
(
n
+
1
)
)
×
(
b
1
1
⋅
2
+
b
2
2
⋅
3
+
.
.
.
+
b
n
n
(
n
+
1
)
)
≤
a
1
b
1
1
⋅
2
+
a
2
b
2
2
⋅
3
+
.
.
.
+
a
n
b
n
n
(
n
+
1
)
.
\left( \frac{a_1}{1 \cdot 2}+\frac{a_2}{2 \cdot 3}+...+\frac{a_n}{n(n+1)} \right) \times \left( \frac{b_1}{1 \cdot 2}+\frac{b_2}{2 \cdot 3}+...+\frac{b_n}{n(n+1)} \right) \leq \frac{a_1b_1}{1 \cdot 2}+\frac{a_2b_2}{2 \cdot 3}+...+\frac{a_nb_n}{n(n+1)}.
(
1
⋅
2
a
1
+
2
⋅
3
a
2
+
...
+
n
(
n
+
1
)
a
n
)
×
(
1
⋅
2
b
1
+
2
⋅
3
b
2
+
...
+
n
(
n
+
1
)
b
n
)
≤
1
⋅
2
a
1
b
1
+
2
⋅
3
a
2
b
2
+
...
+
n
(
n
+
1
)
a
n
b
n
.
Proposed by A. Antropov
Inequality
algebra
inequalities