Example 8 Let $f(a, b, c)=\frac{1}{\sqrt{1+2 a}}+\frac{1}{\sqrt{1+2 b}}+\frac{1}{\sqrt{1+2 c}}$, where $a, b, c>0$ and $abc=1$, find the minimum value of the constant $\lambda$ such that $f(a, b, c)<\lambda$ always holds.
This is not a simple problem. We guess the minimum value of $\lambda$ is 2, that is to prove $\frac{1}{\sqrt{1+2 a}}+$ $\frac{1}{\sqrt{1+2 b}}+\frac{1}{\sqrt{1+2 c}}\frac{\sqrt{a b}}{\sqrt{2+a b}}$,
Since $\frac{a}{1+2 a}+\frac{b}{1+2 b} \geqslant \frac{2 \sqrt{a b}}{\sqrt{(1+2 a)(1+2 b)}}$, and $\frac{2 \sqrt{a b}}{\sqrt{(1+2 a)(1+2 b)}}>$ $\frac{\sqrt{a b}}{\sqrt{2+a b}} \Leftrightarrow 4(2+a b) \geqslant(1+2 a)(1+2 b) \Leftrightarrow a+b \leqslant \frac{7}{2}$.
When $a+b>\frac{7}{2}$, (we can assume $a \leqslant b \leqslant c$), then $b>\frac{7}{4}, c>\frac{7}{4}$, at this time $\frac{1}{\sqrt{1+2 b}}+\frac{1}{\sqrt{1+2 c}}<\frac{2}{\sqrt{1+\frac{7}{2}}}=\frac{2 \sqrt{2}}{3}<1$.
Of course, $\frac{1}{\sqrt{1+2 a}}+\frac{1}{\sqrt{1+2 b}}+\frac{1}{\sqrt{1+2 c}}<1+1=2$.
If we take $a=b=\frac{1}{n}, c=n^{2}$, then
$$
\lim _{n \rightarrow+\infty} f(a, b, c)=\lim _{n \rightarrow+\infty}\left(\frac{1}{\sqrt{1+\frac{2}{n}}}+\frac{1}{\sqrt{1+\frac{2}{n}}}+\frac{1}{\sqrt{1+2 n^{2}}}\right)=2 .
$$

This shows that $\lambda=2$ is the minimum value.