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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 28, 2011 4:44 PM
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On Thu, 28 Jul 2011 16:38:29 -0500, quasi <quasi@null.set> wrote:

>"huhie" wrote:
>>"Kenshin" wrote:

>>> x, y, z : positive real number satisfying x+y+z=3
>>> Prove that (x+1)/(b^2+1) + (b+1)/(c^2+1) + (c+1)/(a^2+1) >=3

>>how do I do that? 4 unknowns and one equation ?

>Clearly Kenshin neglected to check his message before (and
>after) posting it.
>Presumably, the intended problem was as follows:
>Prove that if x,y,z are positive real numbers such that
> x + y + z = 3
> x/(y^2+1) + y/(z^2+1) + z/(x^2+1) >= 3

Oops! (but at least I checked _after_ I posted).

I meant to write ...

Presumably, the intended problem was as follows:

Prove that if x,y,z are positive real numbers such that

x + y + z = 3


(x+1)/(y^2+1) + (y+1)/(z^2+1) + (z+1)/(x^2+1) >= 3


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