`B=(1-1/x^2)(1-1/y^2)`
`B=(1+1/x)(1+1/y). (1-1/x)(1-1/y)`
`B=[(x+1)(y+1)]/(xy). [(x-1)(y-1)]/(xy)`
`B=(xy+x+y+1)/(xy). (xy-x-y+1)/(xy)`
`B=(xy+2)/(xy). (xy)/(xy)`
`B=1+2/(xy)`
Theo BĐT $Cô-si$
`xy\le (x+y)^2/4=1/4`
`⇒B=1+2/(xy)\le 1+2/(1/4)=9`
Dấu `=` xảy ra $\begin{cases}x+y=1\\x=y\end{cases}⇔x=y=\dfrac{1}{2}$
Vậy $Min_B=9⇔x=y=\dfrac{1}{2}$