回復 1# weiye 的帖子
\(\begin{align}
& a>0 \\
& {{b}^{2}}-4ac\le 0,c\ge \frac{{{b}^{2}}}{4a} \\
& \frac{a+b+c}{b-a} \\
& \ge \frac{a+b+\frac{{{b}^{2}}}{4a}}{b-a} \\
& =\frac{1+\frac{b}{a}+\frac{{{b}^{2}}}{4{{a}^{2}}}}{\frac{b}{a}-1} \\
& =\frac{\frac{1}{4}{{\left( \frac{b}{a}-1 \right)}^{2}}+\frac{3}{2}\left( \frac{b}{a}-1 \right)+\frac{9}{4}}{\frac{b}{a}-1} \\
& =\frac{1}{4}\left( \frac{b}{a}-1 \right)+\frac{\frac{9}{4}}{\frac{b}{a}-1}+\frac{3}{2} \\
& \ge \frac{3}{2}+\frac{3}{2} \\
& =3 \\
\end{align}\)
由於\(m<\frac{a+b+c}{b-a}\)恆成立，故\(m<3\)
