$B=(x^2+1)(y^2+1)-(x+4)(x-4)-(y^2-25)$
$B=x^2y^2+x^2+y^2+1-x^2+16-y^2+25$
$B=(x^2y^2+x^2+y^2-x^2-y^2)+(1+16+25)$
$B=x^2y^2+42$
Vì $x^2y^2≥0⇒B=x^2y^2+42≥42$
Dấu `=` xảy ra $⇔x^2y^2=0$
$⇒\left[ \begin{array}{l}x^2=0\\y^2=0\end{array} \right.$
$⇒\left[ \begin{array}{l}x=0\\y=0\end{array} \right.$
$⇒B=42⇔x=0$ hoặc $y=0$