149 Given the quadratic function $f(x)=x^{2}-2 m x+1$. Does there exist a real number $x$, such that for any real numbers $a, b, c$ satisfying $0 \leqslant x \leqslant 1$, $f(a), f(b), f(c)$ can form the lengths of the three sides of a triangle.
Solution: This problem is equivalent to:
For $0 \leqslant x \leqslant 1, f(x)_{\text {min }} \geqslant 0$ and $2 f(x)_{\text {min }}>f(x)_{\text {max }}$.
(1) When $m 0 , } \\
{ 2 f ( 0 ) > f ( 1 ) }
\end{array} \Rightarrow \left\{\begin{array}{l}
1>0, \\
2>2-2 m
\end{array} \Rightarrow m>0 .\right.\right.
$$

When $m0, \\
2\left(1-m^{2}\right)>2-2 m
\end{array} \Rightarrow 00, \\
2\left(1-m^{2}\right)>1
\end{array} \Rightarrow-\frac{\sqrt{2}}{2}1$, $f(1)$ is the minimum, and $f(0)$ is the maximum.
$$
\left\{\begin{array}{l}
f(1)=2-2 m>0, \\
2(2-2 m)>1
\end{array} \Rightarrow m1$ is a contradiction.
In summary, when $0<m<\frac{\sqrt{2}}{2}$, the conditions of the problem are satisfied.