quasi
Posts:
9,076
Registered:
7/15/05
|
|
Re: An cyclic inequality
Posted:
Jul 28, 2011 4:44 PM
|
|
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
>
>then
>
> 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
then
(x+1)/(y^2+1) + (y+1)/(z^2+1) + (z+1)/(x^2+1) >= 3
quasi
|
|
