1. If $f(x)=x^{3}-3 x+m$ takes any three numbers $a, b, c$ in the interval $[0,2]$, there exists a triangle with side lengths $f(a), f(b), f(c)$, then the range of $m$ is $\qquad$
1. $(6,+\infty)$.
Notice that, $f^{\prime}(x)=3(x+1)(x-1)$.
When $x \in[0,1)$, $f^{\prime}(x)0, f(x)$ is increasing on (1,2].
Thus, when $x \in[0,2]$,
$$
\begin{array}{l}
f(x)_{\min }=f(1)=m-2>0, \\
f(2)=2+m>f(0)=m .
\end{array}
$$

By the given condition $2 f(1)>f(2)$.
Therefore, $m>6$.