Example 13 Let $a, b, c$ all be positive numbers, and $a+b+c=\lambda$. If the inequality $\frac{1}{a(1+\lambda b)}+\frac{1}{b(1+\lambda c)}+\frac{1}{c(1+\lambda a)} \geqslant \frac{27}{4}$ always holds, find the range of values for $\lambda$.

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Let's construct the vectors $\boldsymbol{a}=\left(\sqrt{\frac{1}{a(1+\lambda b)}}, \sqrt{\frac{1}{b(1+\lambda c)}}, \sqrt{\frac{1}{c(1+\lambda a)}}\right)$,
$$
\boldsymbol{b}=(\sqrt{a(1+\lambda b)}, \sqrt{b(1+\lambda c)}, \sqrt{c(1+\lambda a)}) .
$$

Then $\boldsymbol{a} \cdot \boldsymbol{b}=3$,
$$
|\boldsymbol{a}| \cdot|\boldsymbol{b}|=\sqrt{\frac{1}{a(1+\lambda b)}+\frac{1}{b(1+\lambda c)}+\frac{1}{c(1+\lambda a)}} \cdot \sqrt{(a+b+c)+\lambda(a b+b c+c a)} .
$$

By $\lambda^{2}=(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \geqslant 3(a b+b c+c a)$,
we get $a b+b c+c a \leqslant \frac{\lambda^{2}}{3}$,
$$
|\boldsymbol{a}| \cdot|\boldsymbol{b}| \leqslant \sqrt{\frac{1}{a(1+\lambda b)}+\frac{1}{b(1+\lambda c)}+\frac{1}{c(1+\lambda a)}} \cdot \sqrt{\lambda+\frac{1}{3} \lambda^{3}},
$$

which implies $\frac{1}{a(1+\lambda b)}+\frac{1}{b(1+\lambda c)}+\frac{1}{c(1+\lambda a)} \geqslant \frac{27}{3 \lambda+\lambda^{3}}$.
Equality holds if and only if $a=b=c$ and $\boldsymbol{a}, \boldsymbol{b}$ are in the same direction, i.e., $a=b=c=\frac{\lambda}{3}$.
Thus, the given inequality holds for all $\lambda$ if and only if $\frac{27}{3 \lambda+\lambda^{3}} \geqslant \frac{27}{4}(\lambda>0) \Leftrightarrow(\lambda-1)\left(\lambda^{2}+\lambda+4\right) \leqslant 0(\lambda>0) \Leftrightarrow 0<\lambda \leqslant 1$.
Therefore, the range of $\lambda$ is the interval $(0,1]$.