6. Let $m>1$, under the constraint conditions
$$
\left\{\begin{array}{l}
y \geqslant x, \\
y \leqslant m x, \\
x+y \leqslant 1
\end{array}\right.
$$

the maximum value of the objective function $z=x+m y$ is less than 2. Then the range of values for $m$ is . $\qquad$
6. $(1,1+\sqrt{2})$.

The objective function is transformed into $y=-\frac{1}{m} x+\frac{z}{m}$.
Given $m>1$, we know $-1<-\frac{1}{m}<0$.
The feasible region represented by the system of inequalities is the shaded area in Figure 2.

According to the geometric meaning of the objective function, the maximum value of the objective function is achieved when the intercept of the line $y=-\frac{1}{m} x+\frac{z}{m}$ on the $y$-axis is the largest.
Clearly, the maximum value is achieved at point $A$.
By solving $\left\{\begin{array}{l}y=m x, \\ x+y=1,\end{array}\right.$, we get the intersection point $A\left(\frac{1}{1+m}, \frac{m}{1+m}\right)$.
Thus, the maximum value of the objective function is
$$
\begin{array}{l}
\frac{1}{1+m}+\frac{m^{2}}{1+m}<2 \Rightarrow m^{2}-2 m-1<0 \\
\Rightarrow 1-\sqrt{2}<m<1+\sqrt{2} .
\end{array}
$$

Therefore, the range of $m$ is $(1,1+\sqrt{2})$.