8 If the inequality $\lg (x y) \leqslant(\lg a) \sqrt{\lg ^{2} x+\lg ^{2} y}$ holds for $x>1, y>1$, then the minimum value of the real number $a$ is $\qquad$
$810^{\sqrt{2}}$. The original inequality can be transformed into $\frac{\lg (x y)}{\sqrt{\lg ^{2} x+\lg ^{2} y}} \leqslant \lg a$. When $x=y$, $\frac{\lg (x y)}{\sqrt{\lg ^{2} x+\lg ^{2} y}}=\sqrt{2}$. Thus, $\lg a \geqslant \sqrt{2}$, which means $a \geqslant 10^{\sqrt{2}}$.