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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

1745.0. "Crux Mathematicorum 1830" by RUSURE::EDP (Always mount a scratch monkey.) Mon Apr 19 1993 16:47

    Proposed by P. Tsaussoglou, Athens, Greece.
    
    If a > b > c > 0 and a^-1 + b^-1 + c^-1 = 1, prove that
    
    	 4      1        1       4
    	--- + ------ + ------ >= -.
    	c^2   (a-b)b   (b-c)c    3
T.RTitleUserPersonal
Name		DateLines
1745.1RUSURE::EDPAlways mount a scratch monkey.Wed Mar 02 1994 09:3631
    Solution by Sefket Arslanagic, Nyborg, Denmark.
    
    From a>b>0 and b>c>0, by the Arithmetic-Mean Geometric-Mean inequality
    we have
    
    	a = (a-b)+b >= 2sqrt((a-b)b)  and  b = (b-c)+c >= 2sqrt((b-c)c),
    
    or
    
    	1/((a-b)b) >= 4/a^2  and  1/((b-c)c) >= 4/b^2.
    
    Now, we have
    
    	4/c^2 + 1/((a-b)b) + 1/((b-c)c) >= 4(1/a^2 + 1/b^2 + 1/c^2)	(1)
    
    Equality in (1) holds if and only if a = 2b = 4c (i.e., a=7, b=7/2, and
    c=7/4 because a^-1 + b^-1 + c^-1 = 1).  Since
    
    	(1/a-1/b)^2 + (1/b-1/c)^2 + (1/c-1/a)^2 > 0
    
    implies that
    
    	1/a^2 + 1/b^2 + 1/c^2 > 1/3 (1/a+1/b+1/c)^2 = 1/3		(2)
    
    (strict inequality holds here because a>b>c>0), it follows from (1)
    that
    
    	4/c^2 + 1/((a-b)b) + 1/((b-c)c) > 4/3,
    
    i.e., the given inequality is correct, but only strict inequality
    holds.