10. Given the sequence $\left\{a_{n}\right\}$ defined by $a_{1}=\frac{2}{3}, a_{n+1}=a_{n}^{2}+a_{n-1}^{2}+\cdots+a_{1}^{2}\left(n \in \mathbf{N}^{*}\right)$. If for any $n$ $\in \mathbf{N}^{*}, \frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}+\cdots+\frac{1}{a_{n}+1}<M$ always holds, find the minimum value of $M$.
10. Solution: From the problem, we know that when $n \geqslant 2$, $a_{n+1}=a_{n}^{2}+a_{n}$. Also, $a_{2}=a_{1}^{2}=\frac{4}{9}=\left(\frac{1}{3}\right)^{2}+\frac{1}{3}$. Let's assume $b_{1}=\frac{1}{3}$, $b_{n}=a_{n} (n \geqslant 2)$, then $b_{n+1}=b_{n}^{2}+b_{n} (n \in \mathbf{N}^{*})$. We will use mathematical induction to prove $\frac{1}{b_{1}+1}+\frac{1}{b_{2}+1}+\cdots+\frac{1}{b_{n}+1}=\frac{3 b_{n+1}-1}{b_{n+1}}$. (*)
When $n=1$, $b_{2}=\frac{4}{9}, \frac{3 b_{2}-1}{b_{2}}=\frac{3}{4}, \frac{1}{b_{1}+1}=\frac{3}{4}$, so when $n=1$, (*) holds.
Assume when $n=k$, $\frac{1}{b_{1}+1}+\frac{1}{b_{2}+1}+\cdots+\frac{1}{b_{k}+1}=\frac{3 b_{k+1}-1}{b_{k+1}}$,
then when $n=k+1$, $\frac{1}{b_{1}+1}+\frac{1}{b_{2}+1}+\cdots+\frac{1}{b_{k}+1}+\frac{1}{1+b_{k+1}}=\frac{3 b_{k+1}-1}{b_{k+1}}+\frac{1}{1+b_{k+1}}=\frac{3 b_{k+1}\left(1+b_{k+1}\right)-1}{b_{k+1}\left(b_{k+1}+1\right)}$.
From $b_{n+1}=b_{n}^{2}+b_{n}$, we know $\frac{3 b_{k+1}\left(1+b_{k+1}\right)-1}{b_{k+1}\left(b_{k+1}+1\right)}=\frac{3 b_{k+2}-1}{b_{k+2}}$, so when $n=k+1$, the conclusion holds,
and (*) is true. Then $\frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}+\cdots+\frac{1}{a_{n}+1}=\frac{1}{b_{1}+1}+\frac{1}{b_{2}+1}+\cdots+\frac{1}{b_{n}+1}+\frac{1}{1+a_{1}}-\frac{1}{1+b_{1}}=\frac{3 b_{n+1}-1}{b_{n+1}}-\frac{3}{20}=\frac{57}{20}-\frac{1}{b_{n+1}}$, given that $b_{n+1}$ is a positive number, and $b_{n+1}-b_{n}=b_{n}^{2} \geqslant b_{1}^{2}=\frac{1}{9}$, so as $n$ approaches infinity, $b_{n+1}$ approaches infinity, then the minimum value of the required $M$ is $\frac{57}{20}$.