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2
Inequality with abc=1
Inequality with abc=1
Source: JBMO Shortlist 2012 A2
February 4, 2015
inequalities
inequalities proposed
Problem Statement
Let
a
a
a
,
b
b
b
,
c
c
c
be positive real numbers such that
a
b
c
=
1
abc=1
ab
c
=
1
. Show that :
1
a
3
+
b
c
+
1
b
3
+
c
a
+
1
c
3
+
a
b
≤
(
a
b
+
b
c
+
c
a
)
2
6
\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab} \leq \frac{ \left (ab+bc+ca \right )^2 }{6}
a
3
+
b
c
1
+
b
3
+
c
a
1
+
c
3
+
ab
1
≤
6
(
ab
+
b
c
+
c
a
)
2
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