Re: rewrite this inequality

From: Robert Israel (israel_at_math.ubc.ca)
Date: 12/09/04 

Date: 9 Dec 2004 18:17:15 GMT

In article <41b836f1$1_2@Usenet.com>,
stupid_mathematician <newton_0102@yahoo-dot-com.no-spam.invalid> wrote:
>Sorry, this inequality is:
>[tex]\frac{1}{a^2(b^2+1)} + \frac{1}{b^2(c^2+1)}\frac{1}{c^2(a^2+1)}
>>= \frac{1}{ab(bc+1)}+\frac{1}{bc(ca+1)}+\frac{1}{ca(ab+1)}[/tex]

I think you left out a "+": as it is the inequality is obviously false for
a=b=c=1. Perhaps you mean

\frac{1}{a^2(b^2+1)} + \frac{1}{b^2(c^2+1)} + \frac{1}{c^2(a^2+1)}
>= \frac{1}{ab(bc+1)}+\frac{1}{bc(ca+1)}+\frac{1}{ca(ab+1)}

for a, b, c > 0. But even then it's false: try a=1, b=1/c when
c is large.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada