4.22 Given that $a, b, c$ are real numbers, and $b+c=6-4a+3a^{2}, c-b=4-4a+a^{2}$. Then the size relationship between $a, b, c$ is
(A) $a<b \leqslant c$.
(B) $b<a \leqslant c$.
(C) $b<c \leqslant a$.
(D) $c<a \leqslant b$.
(China Sichuan Province Junior High School Mathematics Competition, 1992)
[Solution] From the given, $c-b=4-4a+a^{2}=(a-2)^{2} \geqslant 0$, we know $c \geqslant b$. By subtracting the two given equations, we have $2b=2+2a^{2}$, which simplifies to $b=1+a^{2}$.
Also, $b-a=a^{2}+1-a$,
and $a^{2}-a+1>0$ (noting that the coefficient of the quadratic term is positive, and $\Delta=1^{2}-4a$.
Thus, $a<b \leqslant c$.
Therefore, the answer is $(A)$.