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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1664.0. "CMJ Problem 485" by BEING::EDP (Always mount a scratch monkey.) Fri Sep 11 1992 13:56

    Prove that
    
    	i) if a >= b > 1 or 1 > a >= b > 0, then
    		a^b^b*b^a^a >= a^b^a*b^a^b; and
    	ii) if a > 1 > b > 0, then a^b^b*b^a^a <= a^b^a*b^a^b.
    
    (Exponentiation associates on the right, so x^y^z = x^(y^z).)
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