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

From the problem, we know $a>0$, and $b^{2}-4 a c \leqslant 0$, so $c \geqslant \frac{b^{2}}{4 a}$. Let $b-a=t(t \geqslant 0)$, i.e., $b=a+t$. 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., $b=c=4 a$. 
Therefore, $M_{\max }=\frac{1}{3}$.