10. (20 points) Given $f(x)=\frac{2 x}{x+1}$. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=2$, and
$$
a_{n}=\frac{1}{2} f\left(a_{n-1}\right)\left(n \in \mathbf{N}_{+}, n \geqslant 2\right) \text {, }
$$
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove: For all positive integers $n$,
$$
a_{1} a_{2} a_{3}+a_{2} a_{3} a_{4}+\cdots+a_{n} a_{n+1} a_{n+2}<\frac{2}{3} .
$$
10. (1) From the given, we have
$$
\begin{array}{l}
a_{n}=\frac{a_{n-1}}{a_{n-1}+1} \\
\Rightarrow \frac{1}{a_{n}}=1+\frac{1}{a_{n-1}}=\cdots=(n-1)+\frac{1}{a_{1}} \\
\quad=\frac{2 n-1}{2} \\
\Rightarrow a_{n}=\frac{2}{2 n-1} .
\end{array}
$$

When $n=1$, $a_{1}=2$, which also fits the above formula.
Thus, $a_{n}=\frac{2}{2 n-1}\left(n \in \mathbf{N}_{+}\right)$.
(2) For $k=1,2, \cdots$, note that
$$
\begin{array}{l}
a_{k} a_{k+1} a_{k+2}=\frac{8}{(2 k-1)(2 k+1)(2 k+3)} \\
=2\left[\frac{1}{(2 k-1)(2 k+1)}-\frac{1}{(2 k+1)(2 k+3)}\right] .
\end{array}
$$

Therefore, $a_{1} a_{2} a_{3}+a_{2} a_{3} a_{4}+\cdots+a_{n} a_{n+1} a_{n+2}$
$$
=2\left[\frac{1}{1 \times 3}-\frac{1}{(2 n+1)(2 n+3)}\right]<\frac{2}{3} .
$$