7 If for any real number $x$, the function
$$
f(x)=x^{2}-2 x-|x-1-a|-|x-2|+4
$$

is always a non-negative real number, then the minimum value of the real number $a$ is . $\qquad$
(7) - 2 Hint: From the conditions, we have $\left\{\begin{array}{l}f(0)=-|1+a|+2 \geqslant 0, \\ f(1)=-|a|+2 \geqslant 0,\end{array}\right.$ Solving this, we get $-2 \leqslant a \leqslant 1$.
When $a=-2$,
$$
f(x)=x^{2}-2 x-|x+1|-|x-2|+4,
$$

i.e.,
$$
f(x)=\left\{\begin{array}{ll}
x^{2}+3, & x2 .
\end{array}\right.
$$

It is easy to see that for any real number $x$, the value of $f(x)$ is a non-negative real number, thus, $a=-2$ meets the requirement.
Therefore, the minimum value of the real number $a$ is -2.