12. Let the continuous monotonic function $f: \mathbf{R} \rightarrow \mathbf{R}, f(0)=1$, and satisfy the inequality
$$
f(x+y) \geqslant f(x) f(y)-f(x y)+1 .
$$

Find $f(x)$.
12. From inequality (1), we get
$$
f(x+y)-f(x) \geqslant f(x)[f(y)-1]+1-f(x y),
$$

For $y>0$, it becomes
$$
\frac{f(x+y)-f(x)}{y} \geqslant f(x)\left[\frac{f(y)-f(0)}{y-0}\right]-\frac{f(0)-f(x y)}{0-x y} \cdot x \cdot
$$

Let $y \rightarrow 0$, we get
$$
f^{\prime}(x) \geqslant f(x) f^{\prime}(0)-f^{\prime}(0) x,
$$

For $y<0$, it becomes

Let $y \rightarrow 0$, we get
$$
f^{\prime}(x) \leqslant f(x) f^{\prime}(0)-f^{\prime}(0) x,
$$

Therefore, $f^{\prime}(x)=f(x)-x$.
$$
\begin{array}{c}
{\left[\mathrm{e}^{-x} f(x)\right]^{\prime}=\mathrm{e}^{-x} f^{\prime}(x)-\mathrm{e}^{-x} f(x)=-x \mathrm{e}^{-x},} \\
\mathrm{e}^{-x} f(x)=\int\left(-x \mathrm{e}^{-x}\right) \mathrm{d} x, \text { i.e., } \mathrm{e}^{-x} f(x)=x \mathrm{e}^{-x}+\mathrm{e}^{-x}+c
\end{array}
$$

Given $f(0)=1$, we get $c=0$, so $f(x)=x+1$.
Verification: When $f(x)=x+1$, it is easy to verify that $f(x+y)=f(x) f(y)-f(x y)+1$.