25. Let $f: \mathbf{R}^{+} \rightarrow \mathbf{R}^{+}$ be a continuous function, $f(1)=1$, for $\forall x, \in \mathbf{R}^{+}, i=1,2, \cdots, n, n \in \mathbf{N}^{+}$, solve the functional inequality: $f\left(\prod_{i=1}^{n} x_{i}\right) \leqslant \prod_{i=1}^{n} f\left(x_{i}\right)$.
25. Taking $n=2$, we get $f\left(x_{1} x_{2}\right) \leqslant f\left(x_{1}\right)+f\left(x_{2}\right)$,
$$
f\left(x_{1} x_{2}\right)-f\left(x_{1}\right) \leqslant f\left(x_{1}\right)\left[f\left(x_{2}\right)-1\right] .
$$

Noting that $f(1)=1$,
when $x_{2}>1$, dividing both sides of (1) by $x_{2}-1$, we get
$$
\frac{f\left(x_{1} x_{2}\right)-f\left(x_{1}\right)}{x_{1} x_{2}-x_{1}} \cdot x_{1} \leqslant f\left(x_{1}\right) \cdot \frac{f\left(x_{2}\right)-f(1)}{x_{2}-1} .
$$

Letting $x_{2} \rightarrow 1$, we get
$$
x_{1} f^{\prime}\left(x_{1}\right) \leqslant f\left(x_{1}\right) f^{\prime}(1) .
$$

When $x_{2}<1$, dividing both sides of (1) by $x_{2}-1$, we get
$$
\begin{aligned}
\frac{f\left(x_{1} x_{2}\right)-f\left(x_{1}\right)}{x_{1} x_{2}-x_{1}} & \geqslant \frac{f\left(x_{1}\right) \cdot f\left(x_{2}\right)-f(1)}{x_{2}-1}, \\
x_{1} f^{\prime}\left(x_{1}\right) & \geqslant f\left(x_{1}\right) f^{\prime}(1) . \\
x_{1} f^{\prime}\left(x_{1}\right) & =f\left(x_{1}\right) f^{\prime}(1) .
\end{aligned}
$$

Letting $x_{2} \rightarrow 1$, we get
From (2) and (3), we have
that is, $x_{1}\left[\ln f\left(x_{1}\right)\right]^{\prime}=f^{\prime}(1), \ln f\left(x_{1}\right)=f^{\prime}(1) \ln x_{1}$, so $f\left(x_{1}\right)=x_{1}^{\prime(1)}$, hence $f(x)=x^{\prime \prime}$.