proved

(x/(1+xy))^2+(y/(1+yz))^2+(z/(1+zx))^2=(zx/(1+z))^2+(xy/(1+x))^2+(yz/(1+y))^2=1/2 (zx/(1+z))^2+1/2 (xy/(1+x))^2+1/2 (yz/(1+y))^2+1/2 (zx/(1+z))^2+1/2 (xy/(1+x))^2+1/2 (yz/(1+y))^2≥zx/(1+z) xy/(1+x)+yz/(1+y) zx/(1+z)+xy/(1+x) yz/(1+y)=x/(1+z)(1+x) +z/(1+y)(1+z) +y/(1+x)(1+y) ==(x(1+y)+z(1+x)+y(1+z))/(1+z)(1+y)(1+x) =(x+yz+y+zx+z+xy)/(1+z)(1+y)(1+x) =(x+1/x+y+1/y+z+1/z)/(1+z)(1+y)(1+x) =((x-1)^2/x+(y-1)^2/y+(z-1)^2/z+6)/(1+z)(1+y)(1+x)
(1+z)(1+y)(1+x)=(1+z)(1+x+y+xy)=1+x+y+xy+z+zx+zy+1=2+x+1/x+y+1/y+z+1/z=(x-1)^2/x+(y-1)^2/y+(z-1)^2/z+8
(x/(1+xy))^2+(y/(1+yz))^2+(z/(1+zx))^2≥((x-1)^2/x+(y-1)^2/y+(z-1)^2/z+6)/((x-1)^2/x+(y-1)^2/y+(z-1)^2/z+8)
(a+6)/(a+8)=(a+8-2)/(a+8)=1-2/(a+8)≥1- 2/8=3/4 for any a>0

所有跟帖： 

• 妙！（為便於閱讀，第5行，第6行宜另起兩段） -wxcfan123- ♂ (72 bytes) () 08/09/2015 postreply 18:42:07