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Latvia BW TST
2015 Latvia Baltic Way TST
12
sum a^{2014}/(1 + 2 bc) >=3/ (ab+bc+ca) for a,b,c<0 with abc = 1
sum a^{2014}/(1 + 2 bc) >=3/ (ab+bc+ca) for a,b,c<0 with abc = 1
Source: 2015 Latvia BW TST P12
December 16, 2022
algebra
inequalities
Problem Statement
For real positive numbers
a
,
b
,
c
a, b, c
a
,
b
,
c
, the equality
a
b
c
=
1
abc = 1
ab
c
=
1
holds. Prove that
a
2014
1
+
2
b
c
+
b
2014
1
+
2
a
c
+
c
2014
1
+
2
a
b
≥
3
a
b
+
b
c
+
c
a
.
\frac{a^{2014}}{1 + 2 bc}+\frac{b^{2014}}{1 + 2ac}+\frac{c^{2014}}{1 + 2ab} \ge \frac{3}{ab+bc+ca}.
1
+
2
b
c
a
2014
+
1
+
2
a
c
b
2014
+
1
+
2
ab
c
2014
≥
ab
+
b
c
+
c
a
3
.
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