5. For all real numbers $x$, the values of all quadratic functions $f(x) = ax^2 + bx + c$ $(a < b)$ are non-negative real numbers. Then the maximum value of $\frac{b-a}{a+b+c}$ is ( ).
(A) $\frac{1}{3}$
(B) $\frac{1}{2}$
(C) 3
(D) 2
5. (A).

From the problem, we know that $a>0$ and $b^{2}-4 a c \leqslant 0$, which means $c \geqslant \frac{b^{2}}{4 a}$.
Let $b-a=t \quad(t \geqslant 0)$, then $b=a+t$. Therefore,
$$
\begin{array}{l} 
M= \frac{b-a}{a+b+c} \leqslant \frac{t}{a+(a+t)+\frac{(a+t)^{2}}{4 a}} \\
=\frac{4 a t}{(3 a+t)^{2}} \leqslant \frac{4 a t}{(2 \sqrt{3 a t})^{2}}=\frac{1}{3} .
\end{array}
$$

Equality holds if and only if $3 a=t$, and $c=\frac{b^{2}}{4 a}$, i.e., $b=c=4 a$. Hence, $M_{\max }=\frac{1}{3}$.