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2019 JBMO Shortlist
A1
A1
Part of
2019 JBMO Shortlist
Problems
(1)
JBMO Shortlist 2019 A1
Source:
9/12/2020
Real numbers
a
a
a
and
b
b
b
satisfy
a
3
+
b
3
−
6
a
b
=
−
11
a^3+b^3-6ab=-11
a
3
+
b
3
−
6
ab
=
−
11
. Prove that
−
7
3
<
a
+
b
<
−
2
-\frac{7}{3}<a+b<-2
−
3
7
<
a
+
b
<
−
2
. Proposed by Serbia
algebra
inequalities
JBMO