249 Find the range of the real number $\lambda$ such that the inequality
$$
\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}} \leqslant \frac{3}{\sqrt{1+\lambda}}
$$

holds for any positive real numbers $x, y, z$ satisfying $x y z=\lambda^{3}$.
Solution: In equation (1), let $x=y=t, z=\frac{1}{t^{2}}$.
When $t \rightarrow 0$, we have
$$
\begin{array}{l}
\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}} \\
=\frac{2}{\sqrt{1+t}}+\frac{t}{\sqrt{t^{2}+1}} \rightarrow 2
\end{array}
$$

Then $2 \leqslant \frac{3}{\sqrt{1+\lambda}} \Leftrightarrow 00, r=\sqrt{xy}(01$, then $u+\frac{2(\lambda-1)}{\lambda-2} \sqrt{\lambda+1}>0$.
Also, $(u-\sqrt{\lambda+1})^{2} \geqslant 0, \lambda-2<0$, so equation (6) holds.
Therefore, equation (1) holds.
Thus, the range of $\lambda$ is $\left(0, \frac{5}{4}\right]$.