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Taiwan National Olympiad
2005 Taiwan National Olympiad
1
abc=1
abc=1
Source: junior selection tests-Romania 2003
July 31, 2004
inequalities
function
romania
Problem Statement
Let
a
,
b
,
c
a,b,c
a
,
b
,
c
be three positive real numbers such that
a
b
c
=
1
abc=1
ab
c
=
1
. Prove that:
1
+
3
a
+
b
+
c
≥
6
a
b
+
b
c
+
c
a
.
1+\frac{3}{a+b+c}\ge{\frac{6}{ab+bc+ca}} .
1
+
a
+
b
+
c
3
≥
ab
+
b
c
+
c
a
6
.
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