2、 已知正数 $x$、$y$、$z$ 满足 $x^{2}+y^{2}+z^{2}=1$，则 $s=\dfrac{1+z}{xy}+\dfrac{1}{z}$ 的最小值是__
解：$\because$ 正数 $x$，$y$，$z$ 满足 $x^{2}+y^{2}+z^{2}=1$，
$\therefore x^{2}+y^{2}=1-z^{2}$
$\therefore (x+y)^{2}=1-z^{2}+2xy\geq 4xy$
$\therefore xy\leq\dfrac{1-z^{2}}{2}$
$\therefore \dfrac{1}{xy}\geq\dfrac{2}{1-z^{2}}$
$\therefore s=\dfrac{1+z}{xy}+\dfrac{1}{z}\geq\dfrac{2}{1-z}+\dfrac{1}{z}=\left( \dfrac{2}{1-z}+\dfrac{1}{z} \right)(1-z+z)$
$=3+\dfrac{2z}{1-z}+\dfrac{1-z}{z}\geq 3+2\sqrt{2}$