11. Given positive integers $a, b(a \leqslant b)$. The sequence $\left\{f_{n}\right\}$ satisfies:
$$
f_{1}=a, f_{2}=b, f_{n+2}=f_{n+1}+f_{n}(n=1,2, \cdots) .
$$

If for any positive integer $n$, we have $\left(\sum_{k=1}^{n} f_{k}\right)^{2} \leqslant \lambda \cdot f_{n} f_{n+1}$, find the minimum value of the real number $\lambda$.
Lemma 1: For any $n \in \mathbf{N}^{*}$, we have $f_{n+2}=\sum_{k=1}^{n} f_{k}+f_{2}$.
By the summation method, we get $f_{n+2}=\sum_{k=2}^{n+1}\left(f_{k+1}-f_{k}\right)+f_{2}=\sum_{k=2}^{n+1} f_{k-1}+f_{2}=\sum_{k=1}^{n} f_{k}+f_{2}$.
Lemma 2: Let $\mu=a^{2}+a b-b^{2}$, then for any $n \in \mathbf{N}^{*}$, we have $f_{n} f_{n+2}+(-1)^{n} \mu=f_{n+1}^{2}$. For $n=1$, $f_{1} f_{3}-\mu=a(a+b)-\left(a^{2}+a b-b^{2}\right)=b^{2}=f_{2}^{2}$, the conclusion holds; assuming the conclusion holds for $n=k$, i.e., $f_{k} f_{k+2}+(-1)^{k} \mu=f_{k+1}^{2}$.
Then for $n=k+1$, $f_{k+1} f_{k+3}+(-1)^{k+1} \mu=f_{k+1}\left(f_{k+1}+f_{k+2}\right)-(-1)^{k} \mu$ $=f_{k+1}^{2}+f_{k+1} f_{k+2}-\left(f_{k+1}^{2}-f_{k} f_{k+2}\right)=f_{k+1} f_{k+2}+f_{k} f_{k+2}=f_{k+2}^{2}$, the conclusion also holds. By the principle of mathematical induction, for any $n \in \mathbf{N}^{*}$, we have $f_{n} f_{n+2}+(-1)^{n} \mu=f_{n+1}^{2}$.
Lemma 3: $\lim _{n \rightarrow+\infty} \frac{f_{n}}{f_{n+1}}=\frac{\sqrt{5}-1}{2}$.
First, by $\left|\frac{f_{n}}{f_{n+1}}-\frac{f_{n+1}}{f_{n+2}}\right|=\left|\frac{f_{n} f_{n+2}-f_{n+1}^{2}}{f_{n+1} f_{n+2}}\right|=\left|\frac{(-1)^{n} \mu}{f_{n+1} f_{n+2}}\right| \rightarrow 0(n \rightarrow+\infty)$
we know that $\lim _{n \rightarrow+\infty} \frac{f_{n}}{f_{n+1}}$ exists, let its value be $c$, it is easy to see that $0<c \leqslant 1$.
$$
\begin{array}{l}
f_{n+2}=f_{n+1}+f_{n} \Rightarrow \frac{f_{n+2}}{f_{n+1}}=1+\frac{f_{n}}{f_{n+1}} \Rightarrow \lim _{n \rightarrow+\infty} \frac{f_{n+2}}{f_{n+1}}=1+\lim _{n \rightarrow+\infty} \frac{f_{n}}{f_{n+1}} \\
\Rightarrow \frac{1}{c}=1+c \Rightarrow c^{2}+c-1=0 \Rightarrow c=\frac{\sqrt{5}-1}{2} .
\end{array}
$$

Returning to the original problem, let $T_{n}=\frac{\left(\sum_{k=1}^{n} f_{k}\right)^{2}}{f_{n} f_{n+1}}, n=1,2, \cdots$, then
$T_{n+1}-T_{n}=\frac{\left(\sum_{k=1}^{n+1} f_{k}\right)^{2}}{f_{n+1} f_{n+2}}-\frac{\left(\sum_{k=1}^{n} f_{k}\right)^{2}}{f_{n} f_{n+1}}=\frac{f_{n}\left(\sum_{k=1}^{n} f_{k}+f_{n+1}\right)^{2}-f_{n+2}\left(\sum_{k=1}^{n} f_{k}\right)^{2}}{f_{n} f_{n+1} f_{n+2}}$
$=\frac{f_{n}\left[\left(\sum_{k=1}^{n} f_{k}\right)^{2}+2 f_{n+1} \sum_{k=1}^{n} f_{k}+f_{n+1}^{2}\right]-f_{n+2}\left(\sum_{k=1}^{n} f_{k}\right)^{2}}{f_{n} f_{n+1} f_{n+2}}$
$=\frac{\left(f_{n}-f_{n+2}\right)\left(\sum_{k=1}^{n} f_{k}\right)^{2}+2 f_{n} f_{n+1} \sum_{k=1}^{n} f_{k}+f_{n} f_{n+1}^{2}}{f_{n} f_{n+1} f_{n+2}}$
$=\frac{-f_{n+1}\left(f_{n+2}-f_{2}\right)^{2}+2 f_{n} f_{n+1}\left(f_{n+2}-f_{2}\right)+f_{n} f_{n+1}^{2}}{f_{n} f_{n+1} f_{n+2}}$
$=\frac{-\left(f_{n+2}-f_{2}\right)^{2}+2 f_{n}\left(f_{n+2}-f_{2}\right)+f_{n} f_{n+1}}{f_{n} f_{n+2}}$
$=\frac{2 f_{n+1} f_{2}-f_{2}^{2}+2 f_{n} f_{n+2}+f_{n} f_{n+1}-f_{n+2}^{2}}{f_{n} f_{n+2}}=\frac{2 f_{n+1} f_{2}-f_{2}^{2}+f_{n} f_{n+2}-f_{n+1}^{2}}{f_{n} f_{n+2}}$
$=\frac{2 b f_{n+1}-b^{2}-(-1)^{n} \mu}{f_{n} f_{n+2}}$, noting that $f_{n+1} \geqslant b,-b^{2} \leqslant \mu=a^{2}+a b-b^{2} \leqslant b^{2}$,
then $2 b f_{n+1}-b^{2}-(-1)^{n} \mu \geqslant 2 b^{2}-b^{2}-b^{2}=0 \Rightarrow T_{n+1} \geqslant T_{n}$.
On the other hand, $\lim _{n \rightarrow+\infty} T_{n}=\lim _{n \rightarrow+\infty} \frac{\left(\sum_{k=1}^{n} f_{k}\right)^{2}}{f_{n} f_{n+1}}=\lim _{n \rightarrow+\infty} \frac{\left(f_{n+2}-f_{2}\right)^{2}}{f_{n} f_{n+1}}$
$=\lim _{n \rightarrow+\infty} \frac{\left(f_{n}+f_{n+1}-f_{2}\right)^{2}}{f_{n} f_{n+1}}=\lim _{n \rightarrow+\infty} \frac{\left(f_{n}+f_{n+1}\right)^{2}}{f_{n} f_{n+1}}=\lim _{n \rightarrow+\infty} \frac{f_{n}^{2}+f_{n+1}^{2}+2 f_{n} f_{n+1}}{f_{n} f_{n+1}}$
$=\lim _{n \rightarrow+\infty} \frac{f_{n}}{f_{n+1}}+\lim _{n \rightarrow+\infty} \frac{f_{n+1}}{f_{n}}+2=\frac{\sqrt{5}-1}{2}+\frac{2}{\sqrt{5}-1}+2=2+\sqrt{5}$.
In summary, the supremum of $T_{n}$ is $2+\sqrt{5}$, so the minimum value of the real number $\lambda$ is $2+\sqrt{5}$.