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Topic: An cyclic inequality
Replies: 6   Last Post: Jul 29, 2011 5:01 PM

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Posts: 11,740
Registered: 7/15/05
Re: An cyclic inequality
Posted: Jul 29, 2011 3:44 PM
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On Fri, 29 Jul 2011 15:10:02 -0400, Dan Hoey <haoyuep@aol.com> wrote:

>On 7/28/11 5:42 PM, quasi wrote:

>> Presumably, the intended problem was as follows:
>> Prove that if x,y,z are positive real numbers such that
>> x + y + z = 3
>> then
>> (x+1)/(y^2+1) + (y+1)/(z^2+1) + (z+1)/(x^2+1) >= 3

>It still fails when x=y=z=1.

The symbol >= means "greater than or equal to".

Thus the claimed inequality holds for x = y = z = 1.


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