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Tournament Of Towns
1998 Tournament Of Towns
1
TOT 1998 Spring AS1 sum a^3/a^2+ab+b^2 >=(a+b+c)/3
TOT 1998 Spring AS1 sum a^3/a^2+ab+b^2 >=(a+b+c)/3
Source:
May 11, 2020
inequalities
algebra
Problem Statement
Prove that
a
3
a
2
+
a
b
+
b
2
+
b
3
b
2
+
b
c
+
c
2
+
c
3
c
2
+
c
a
+
a
2
≥
a
+
b
+
c
3
\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\geq \frac{a+b+c}{3}
a
2
+
ab
+
b
2
a
3
+
b
2
+
b
c
+
c
2
b
3
+
c
2
+
c
a
+
a
2
c
3
≥
3
a
+
b
+
c
for positive reals
a
,
b
,
c
a,b,c
a
,
b
,
c
(S Tokarev)
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