1. Find all real functions $f, g, h: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$, we have
$$
\begin{array}{l}
(x-y) f(x)+h(x)-x y+y^{2} \leqslant h(y) \\
\leqslant(x-y) g(x)+h(x)-x y+y^{2} .
\end{array}
$$
(53rd Romanian Mathematical Olympiad (First Round))
Solution: From equation (1), we have
$$
(x-y) f(x) \leqslant (x-y) g(x).
$$

It is easy to see that $f(x) = g(x)$ for all real numbers $x$. Therefore, we have
$$
(x-y) f(x) + h(x) - x y + y^{2} = h(y).
$$

Let $x = 0$, we get $h(y) = y^{2} - f(0) y + h(0)$, which means $h$ is a quadratic function. Define $f(0) = a, h(0) = b$, and substitute $h(y) = y^{2} - a y + b$, we have
$$
(x-y) f(x) + x^{2} - a x + b - x y + y^{2} = y^{2} - a y + b,
$$

which simplifies to $(x-y) f(x) + x(x-y) - (x-y) a = 0$.
Since $x, y$ are arbitrary real numbers, we have $f(x) = -x + a$.
Upon verification, $f(x) = g(x) = -x + a, h(x) = x^{2} - a x + b$ satisfy the conditions, where $a, b$ are real constants, and $x \in \mathbf{R}$.