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




Re: An cyclic inequality
Posted:
Jul 29, 2011 5:01 PM


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



