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2004 IMO Shortlist
5
ab+bc+ca = 1
ab+bc+ca = 1
Source: IMO ShortList 2004, algebra problem 5
March 22, 2005
inequalities
IMO Shortlist
Hi
Problem Statement
If
a
a
a
,
b
b
b
,
c
c
c
are three positive real numbers such that
a
b
+
b
c
+
c
a
=
1
ab+bc+ca = 1
ab
+
b
c
+
c
a
=
1
, prove that
1
a
+
6
b
3
+
1
b
+
6
c
3
+
1
c
+
6
a
3
≤
1
a
b
c
.
\sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}.
3
a
1
+
6
b
+
3
b
1
+
6
c
+
3
c
1
+
6
a
≤
ab
c
1
.
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