Example 2 Let $f(x)$ be a polynomial function defined on the real number field, $f(0)=1$ and for any $x \cdot y \in \mathbf{R}$, it satisfies
$$
f(x-y)+f(x) \geqslant\left(2 x^{2}-2 x y+y^{2}\right)+2 x-y+2
$$

Try to find $f(x)$.
Let $y=x$ be substituted into equation (1), then we have
$$
f(x-x)+f(x)=f(0)+f(x) \geqslant x^{2}+x+2,
$$

which means
$$
1+f(x) \geqslant x^{2}+x+2
$$

Therefore,
$$
f(x) \geqslant x^{2}+x+1, \forall x \in \mathbf{R} .
$$

To verify, $f(x-y)+f(x) \geqslant(x-y)^{2}+(x-y)+1+\left(x^{2}+x+1\right)=\left(2 x^{2}-2 x y+y^{2}\right)+$ $2 x-y+2$, so $f(x) \geqslant x^{2}+x+1$ satisfies (1).