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2
3-var ineq, a+b+c=1
3-var ineq, a+b+c=1
Source: Belarus 2014
February 22, 2020
Inequality
algebra
Belarus
Problem Statement
Let
a
,
b
,
c
a,b,c
a
,
b
,
c
be positive real numbers such that
a
+
b
+
c
=
1
a+b+c=1
a
+
b
+
c
=
1
. Prove that
a
2
(
b
+
c
)
3
+
b
2
(
c
+
a
)
3
+
c
2
(
a
+
b
)
3
≥
9
8
\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3}\geq \frac98
(
b
+
c
)
3
a
2
+
(
c
+
a
)
3
b
2
+
(
a
+
b
)
3
c
2
≥
8
9
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