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

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Dan Hoey

Posts: 172
Registered: 12/6/04
Re: An cyclic inequality
Posted: Jul 29, 2011 5:01 PM
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On 7/29/11 4:34 PM, quasi wrote:
> 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.
>
> quasi


Oops, my arithmetic took a hiccup. Of course you're right.

Dan




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