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Gazeta Matematică 6-7-8/2017
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Gazeta matematică 2017
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From Bitnami MediaWiki
Revision as of 10:58, 2 November 2024 by
Andrei.Horvat
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Gazeta Matematică 6-7-8/2017
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27401
(Radu Pop)
Fie
n
∈
N
{\displaystyle n\in \mathbb {N} }
. Să se arate că
(
n
+
1
)
a
b
(
a
+
b
)
−
(
4
n
3
+
13
n
+
10
)
a
b
+
(
n
+
2
)
3
(
a
+
b
)
≥
(
n
+
2
)
3
,
{\displaystyle (n+1)ab(a+b)-(4n^{3}+13n+10)ab+(n+2)^{3}(a+b)\geq (n+2)^{3},}
oricare ar fi
a
,
b
∈
[
1
,
∞
)
{\displaystyle a,b\in [1,\infty )}
Toggle limited content width
. Să se arate că
