Four. (20 points) Given the function $f_{n}(x)=n^{2} x^{2}(1-$ $x)^{n}, x \in[0,1], n \in \mathbf{N}_{+}$. If the maximum value of $f_{n}(x)$ is denoted as $a_{n}$, try to find the minimum term of the sequence $\left\{a_{n}\right\}$.
It is known that $f_{n}(0)=f_{n}(1)=0$.
When $x \in(0,1)$,
$$
\begin{array}{l}
f_{n}(x)=n^{2} x^{2}(1-x)^{n} \\
=4 \cdot \frac{n x}{2} \cdot \frac{n x}{2} \cdot \underbrace{(1-x)(1-x) \cdots(1-x)}_{n \uparrow} \\
\leqslant 4\left[\frac{\frac{n x}{2}+\frac{n x}{2}+(1-x) n}{n+2}\right]^{n+2}=4\left(\frac{n}{n+2}\right)^{n+2}
\end{array}
$$

The maximum value of $f_{n}(x)$ is achieved when $\frac{n x}{2}=1-x$, i.e., $x=\frac{2}{n+2}$. Thus, $f_{n}(x)$ reaches its maximum value $4\left(\frac{n}{n+2}\right)^{n+2}$. Therefore, $a_{n}=4\left(\frac{n}{n+2}\right)^{n+2}$. Hence,
$$
\begin{array}{l}
\frac{a_{n+1}}{a_{n}}=\frac{4\left(\frac{n+1}{n+3}\right)^{n+3}}{4\left(\frac{n}{n+2}\right)^{n+2}}=\frac{n+1}{n+3} \cdot\left[\frac{(n+1)(n+2)}{n(n+3)}\right]^{n+2} \\
=\frac{n+1}{n+3} \cdot\left[1+\frac{2}{n(n+3)}\right]^{n+2} \\
>\frac{n+1}{n+3} \cdot\left[1+\mathrm{C}_{n+2}^{1} \cdot \frac{2}{n(n+3)}\right] \\
=\frac{n+1}{n+3} \cdot\left[1+\frac{2(n+2)}{n(n+3)}\right] \\
=\frac{n+1}{n+3}+\frac{2}{n+3} \cdot \frac{(n+1)(n+2)}{n(n+3)} \\
>\frac{n+1}{n+3}+\frac{2}{n+3}=1 .
\end{array}
$$

Therefore, $a_{n+1}>a_{n}$, so the sequence $\left\{a_{n}\right\}$ is an increasing sequence.
Thus, the smallest term of the sequence $\left\{a_{n} \mid\right.$ should be $a_{1}=\frac{4}{27}$.