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



quasi
Posts:
12,067
Registered:
7/15/05


Re: An cyclic inequality
Posted:
Jul 29, 2011 3:44 PM


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



