1. If the inequality $\sqrt{x\left(x^{2}+8\right)(8-x)}<\lambda(x+$ 1) holds for all real numbers $x \in(0,2)$, then the range of the real number $\lambda$ is $\qquad$ .
1. $[4,+\infty)$.

Solution: $\frac{\sqrt{x\left(x^{2}+8\right)(8-x)}}{x+1} \leqslant \frac{\left(x^{2}+8\right)+(8 x-8)}{2(x+1)}=$
4, with equality holding if and only if $x=2$.
Therefore, $\frac{\sqrt{x\left(x^{2}+8\right)(8-x)}}{x+1}<4$, hence $\lambda \geqslant 4$.