quasi
Posts:
10,576
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

