10. Given that $\lambda$ is a positive real number, the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=\frac{1}{3}, a_{n+1}=\frac{2 a_{n}^{2}}{4 a_{n}^{2}-2 a_{n}+1}, n=1,2,3, \cdots$. If for any $m \in \mathbf{N}^{*}$, we have $\sum_{k=1}^{m} a_{k}<\lambda$, find the minimum value of $\lambda$.
Given
\[a_{n+1}=\frac{2 a_{n}^{2}}{4 a_{n}^{2}-2 a_{n}+1} \Rightarrow \frac{1}{a_{n+1}}=2-\frac{1}{a_{n}}+\frac{1}{2 a_{n}^{2}},\]
let \(b_{1}=3, b_{n}=\frac{1}{a_{n}}\). When \(b_{n}>2\), we have
\[b_{n+1}-2=\frac{b_{n}^{2}}{2}-b_{n}=\frac{b_{n}\left(b_{n}-2\right)}{2}>0 \Rightarrow b_{n+1}>2.\]
By the principle of mathematical induction, \(b_{n}>2\) for all \(n \in \mathbf{N}^{*}\).

Also,
\[b_{n+1}-b_{n}=\frac{b_{n}^{2}}{2}-2 b_{n}+2=\frac{1}{2}\left(b_{n}-2\right)^{2}>0 \Rightarrow\]
the sequence \(\left\{b_{n}\right\}\) is monotonically increasing. Therefore,
\[b_{n+1}-2=\frac{b_{n}\left(b_{n}-2\right)}{2} \geqslant \frac{3}{2}\left(b_{n}-2\right) \Rightarrow b_{n}-2 \geqslant\left(b_{1}-2\right)\left(\frac{3}{2}\right)^{n-1}=\left(\frac{3}{2}\right)^{n-1} \Rightarrow b_{n} \geqslant 2+\left(\frac{3}{2}\right)^{n-1}.\]

Thus, as \(n \rightarrow+\infty\), \(b_{n} \rightarrow+\infty\).
On the other hand,
\[\frac{1}{b_{n+1}-2}=\frac{2}{b_{n}\left(b_{n}-2\right)}=\frac{1}{b_{n}-2}-\frac{1}{b_{n}}\]
\[\Rightarrow \frac{1}{b_{n}}=\frac{1}{b_{n}-2}-\frac{1}{b_{n+1}-2} \Rightarrow \sum_{k=1}^{m} a_{k}=\sum_{k=1}^{m} \frac{1}{b_{k}}=\frac{1}{b_{1}-2}-\frac{1}{b_{m+1}-2}=1-\frac{1}{b_{m+1}-2}.\]

From the above, we have
\[1-\left(\frac{2}{3}\right)^{m} \leqslant \sum_{k=1}^{m} a_{k}=1-\frac{1}{b_{m+1}-2}<1,\]
so as \(m \rightarrow+\infty\), the supremum of \(\sum_{k=1}^{m} a_{k}\) is 1. Therefore, the minimum value of \(\lambda\) is 1.