\(B=\left(1-\dfrac{1}{x^2}\right)\left(1-\dfrac{1}{y^2}\right)\)
\(=\left(1-\dfrac{1}{x}\right)\left(1+\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\left(1+\dfrac{1}{y}\right)\)
\(=\dfrac{x-1}{x}\cdot\dfrac{y-1}{y}\cdot\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\)
\(=\dfrac{x-\left(x+y\right)}{x}\cdot\dfrac{y-\left(x+y\right)}{y}\cdot\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\)
\(=\dfrac{\left(-y\right)\left(-x\right)}{xy}\cdot\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\)
\(=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\)
\(=1+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{xy}\ge1+\dfrac{4}{x+y}+\dfrac{1}{\dfrac{\left(x+y\right)^2}{4}}=1+\dfrac{4}{1}+\dfrac{1}{\dfrac{1}{4}}=9\)
Vậy \(B_{min}=9\Leftrightarrow x=y=\dfrac{1}{2}\)
