10. Find the smallest positive integer $k$, such that for any $x \in [0,1]$ and $n \in \mathbb{N}_{+}$, the inequality $x^{k}(1-x)^{n}<\frac{1}{(1+n)^{3}}$

always holds.
10. Analysis: By the AM-GM inequality, $x^{k}(1-x)^{n}=\left(\frac{n}{k}\right)^{n} x^{k}\left(\frac{k-k x}{n}\right)^{n} \leq\left(\frac{n}{k}\right)^{n} \cdot\left(\frac{k}{n+k}\right)^{n+k}=\frac{k^{k} n^{n}}{(n+k)^{n+k}}$, so we only need to find the smallest positive integer $k$ such that $\frac{k^{k} n^{n}}{(n+k)^{n+k}}<\frac{1}{(1+n)^{3}}$ for any $n \in N_{+}$. When $n=k=3$, both sides of the inequality are equal, so it does not hold, hence $k \geq 4$. We now prove that $k=4$ holds. That is, we need to prove $4^{4} \cdot n^{n} \cdot(1+n)^{3}<(n+4)^{n+4}$. By the AM-GM inequality,
$$
4^{4} \cdot n^{n} \cdot(1+n)^{3}=8 \cdot(2 n)^{2} \cdot n^{n-2} \cdot(2+2 n)^{3}<\left(\frac{8+4 n+n^{2}-2 n+6+6 n}{n+4}\right)^{n+4}<(n+4)^{n+4}
$$

Proof completed.