7. If the quadratic function $f(x)=a x^{2}+b x+c\left(a, b, c \in \mathbf{R}_{+}\right)$ has zeros, then $\min \left\{\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\right\}$'s maximum value is $\qquad$
7. $\frac{5}{4}$.

Assume $b=2$, then from the quadratic function $f(x)=a x^{2}+b x+c$ having zeros, we get $a c \leqslant 1$. At this point,
$$
\min \left\{\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\right\}=\min \left\{\frac{a+c}{2}, \frac{c+2}{a}, \frac{a+2}{c}\right\}.
$$

Noting the symmetry of $a, c$, assume $a \leqslant c$, so $0<a \leqslant 1$, and $\frac{c+2}{a} \geqslant \frac{a+2}{c}$, thus
$$
\min \left\{\frac{a+c}{2}, \frac{c+2}{a}, \frac{a+2}{c}\right\}=\min \left\{\frac{a+c}{2}, \frac{a+2}{c}\right\}.
$$

Furthermore, when $c \geqslant 2$, we have
$$
\min \left\{\frac{a+c}{2}, \frac{a+2}{c}\right\}=\frac{a+2}{c} \leqslant \frac{\frac{1}{2}+2}{2}=\frac{5}{4}.
$$

When $a \leqslant c<2$, we have
$$
\min \left\{\frac{a+c}{2}, \frac{a+2}{c}\right\}=\frac{a+2}{c}.
$$

In this case, if $0<a \leqslant \frac{1}{2}$, then
$$
\frac{a+c}{2}<\frac{\frac{1}{2}+2}{2}=\frac{5}{4}.
$$

If $\frac{1}{2}<a \leqslant 1$, then
$$
\frac{a+c}{2} \leqslant \frac{a+\frac{1}{a}}{2}<\frac{5}{4}.
$$

In summary, the maximum value sought is $\frac{5}{4}$. This value is achieved when $a: b: c=1: 4: 4$ or $a: b: c=4: 4: 1$.