$$
\begin{array}{l}
\text { Three. (25 points) Given the quadratic function } \\
y=a x^{2}+b x+c(a<b)
\end{array}
$$

the graph is always above or on the $x$-axis, and $m<\frac{a+b+c}{b-a}$ always holds. Find the range of values for $m$.
Three, it is known that $y=a\left(x+\frac{b}{2 a}\right)^{2}+c-\frac{b^{2}}{4 a} \geqslant 0$ always holds.
Therefore, $a>0, c \geqslant \frac{b^{2}}{4 a}$.
From $m1$. Thus,
$$
\begin{array}{l}
\frac{a+b+c}{b-a} \geqslant \frac{a+b+\frac{b^{2}}{4 a}}{b-a}=\frac{4 a^{2}+4 a b+b^{2}}{4 a(b-a)} \\
=\frac{\left(\frac{b}{a}\right)^{2}+4 \times \frac{b}{a}+4}{4\left(\frac{b}{a}-1\right)}=\frac{t^{2}+4 t+4}{4(t-1)} \\
=\frac{\left(t^{2}-1\right)+4(t-1)+9}{4(t-1)} \\
=\frac{1}{4}\left[(t+1)+4+\frac{9}{t-1}\right] \\
=\frac{1}{4}\left(t-1+\frac{9}{t-1}+6\right) \\
\geqslant \frac{1}{4}(6+6)=3 .
\end{array}
$$

When and only when $t=4$, i.e., $\frac{b}{a}=4, b^{2}=4 a c$, the equality holds, so, $\left(\frac{a+b+c}{b-a}\right)_{\min }=3$.
Therefore, $m<3$.