For any
positive real numbers $a_{1}, a_{2}, \cdots, a_{5}$, if $\sum_{i=1}^{5} \frac{a_{i}}{\sqrt{a_{i}^{2}+2^{i-1} a_{i+1} a_{i+2}}}$ $\geqslant \lambda$, find the maximum value of $\lambda$.
Let $t_{i}=2^{i-1} \frac{a_{i+1} a_{i+2}}{a_{i}^{2}}$. Then $\prod_{i=1}^{5} t_{i}=2^{10}$.
Thus, $\sum_{i=1}^{5} \frac{1}{\sqrt{1+t_{i}}} \geqslant \lambda$.
First, we prove a lemma.
Lemma If positive real numbers $x y=c$ (a constant), then
$$
f=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}},
$$

If $x y=1$, then $f$ reaches its minimum value when $x=y$.
(i) If $x y<1$, then $g(t)$ opens upwards, and if
axis $\frac{1}{1-x y} \leqslant \frac{1}{2} \cdot \frac{1}{1+\sqrt{x y}} \Leftrightarrow x y \leqslant \frac{1}{9}$, $x=y$ reaches the minimum value.
(ii) If $x y=1$, then $g(t)$ opens upwards, and if
axis $\frac{1}{1-x y}=\frac{1}{2} \cdot \frac{1}{1+\sqrt{x y}} \Leftrightarrow x y=1$, $x=y$ reaches the minimum value.
(iii) If $x y>1$, then $g(t)$ opens downwards, and if
axis $\frac{1}{x y-1}>\frac{1}{2} \cdot \frac{1}{1+\sqrt{x y}} \Leftrightarrow x y>9$, $x=y$ reaches the minimum value.
(iv) If $x y>1$, then $g(t)$ opens downwards, and if
axis $\frac{1}{x y-1} \leqslant \frac{1}{2} \cdot \frac{1}{1+\sqrt{x y}} \Leftrightarrow x y \geqslant 9$, $x=y$ reaches the minimum value.

In summary, the original expression is greater than 1, and when $a_{1}=n^{6}, a_{2}=1, a_{3}=n, a_{4}=a_{5}=n^{2}, n \rightarrow+\infty$, the limit of the original expression is 1.
Therefore, the maximum value of $\lambda$ is 1.