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1999 Greece Junior Math Olympiad
1
a^{2000}+b^{2000}=a^{1998}+b^{1998} (Greece Junior 1999 p1)
a^{2000}+b^{2000}=a^{1998}+b^{1998} (Greece Junior 1999 p1)
Source:
March 17, 2020
inequalities
algebra
Problem Statement
Show that if
a
,
b
a,b
a
,
b
are positive real numbers such that
a
2000
+
b
2000
=
a
1998
+
b
1998
a^{2000}+b^{2000}=a^{1998}+b^{1998}
a
2000
+
b
2000
=
a
1998
+
b
1998
then a^2+ b^2 \le 2.
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