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Balkan MO
2010 Balkan MO
1
Cyclic sum a^2b(b-c) / (a+b) >= 0
Cyclic sum a^2b(b-c) / (a+b) >= 0
Source: Balkan MO 2010, Problem 1
May 4, 2010
inequalities
rearrangement inequality
inequalities proposed
Balkan
Problem Statement
Let
a
,
b
a,b
a
,
b
and
c
c
c
be positive real numbers. Prove that
a
2
b
(
b
−
c
)
a
+
b
+
b
2
c
(
c
−
a
)
b
+
c
+
c
2
a
(
a
−
b
)
c
+
a
≥
0.
\frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0.
a
+
b
a
2
b
(
b
−
c
)
+
b
+
c
b
2
c
(
c
−
a
)
+
c
+
a
c
2
a
(
a
−
b
)
≥
0.
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