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Belarus Team Selection Test
2009 Belarus Team Selection Test
1
sum 1/(a+b)b >= 9/2(ab+bc+ca) for a,b,c>0
sum 1/(a+b)b >= 9/2(ab+bc+ca) for a,b,c>0
Source: 2009 Belarus TST 2.1
November 8, 2020
inequalities
algebra
Belarus
Problem Statement
Prove that any positive real numbers a,b,c satisfy the inequlaity
1
(
a
+
b
)
b
+
1
(
b
+
c
)
c
+
1
(
c
+
a
)
a
≥
9
2
(
a
b
+
b
c
+
c
a
)
\frac{1}{(a+b)b}+\frac{1}{(b+c)c}+\frac{1}{(c+a)a}\ge \frac{9}{2(ab+bc+ca)}
(
a
+
b
)
b
1
+
(
b
+
c
)
c
1
+
(
c
+
a
)
a
1
≥
2
(
ab
+
b
c
+
c
a
)
9
I.Voronovich
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