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National and Regional Contests
Belarus Contests
Belarus Team Selection Test
2011 Belarus Team Selection Test
2
sum a^3 bc/(b+c) >= 1/6 (sum ab)^2 if sum a/(b+c)=1+1/6 (sum a/c) for a,b,c>0
sum a^3 bc/(b+c) >= 1/6 (sum ab)^2 if sum a/(b+c)=1+1/6 (sum a/c) for a,b,c>0
Source: 2011 Belarus TST 5.2
November 7, 2020
algebra
inequalities
Problem Statement
Positive real
a
,
b
,
c
a,b,c
a
,
b
,
c
satisfy the condition
a
b
+
c
+
b
a
+
c
+
c
a
+
b
=
1
+
1
6
(
a
c
+
b
a
+
c
b
)
\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1+\frac{1}{6}\left( \frac{a}{c}+\frac{b}{a}+\frac{c}{b} \right)
b
+
c
a
+
a
+
c
b
+
a
+
b
c
=
1
+
6
1
(
c
a
+
a
b
+
b
c
)
Prove that
a
3
b
c
b
+
c
+
b
3
c
a
a
+
c
+
c
3
a
b
a
+
b
≥
1
6
(
a
b
+
b
c
+
c
a
)
2
\frac{a^3bc}{b+c}+\frac{b^3ca}{a+c}+\frac{c^3ab}{a+b}\ge \frac{1}{6}(ab+bc+ca)^2
b
+
c
a
3
b
c
+
a
+
c
b
3
c
a
+
a
+
b
c
3
ab
≥
6
1
(
ab
+
b
c
+
c
a
)
2
I.Voronovich
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