Example 2 Find the largest positive number $\lambda$, such that for any real numbers $a, b$, we have $\lambda a^{2} b^{2}(a+b)^{2} \leqslant\left(a^{2}+a b+\right.$ $\left.b^{2}\right)^{3}$.
(Mathematics in Middle School, Issue 4, 2003 Olympiad Training Problem)
When $a b>0$,
$$
\begin{aligned}
\frac{a^{2}+a b+b^{2}}{3} & =\frac{a^{2}+b^{2}+4 a b+3\left(a^{2}+b^{2}\right)}{12} \\
& \geqslant \frac{a^{2}+b^{2}+4 a b+6 a b}{12}=\frac{a^{2}+10 a b+b^{2}}{12} \\
& =\frac{1}{3}\left[\frac{1}{4}(a+b)^{2}+a b+a b\right] \geqslant \sqrt[3]{\frac{a^{2} b^{2}(a+b)^{2}}{4}},
\end{aligned}
$$

The equality holds if and only if $\left\{\begin{array}{l}a=b, \\ \frac{1}{4}(a+b)^{2}=a b,\end{array}\right.$ i.e., $a=b$;
When $a b \leqslant 0$,
$$
\begin{aligned}
\frac{a^{2}+a b+b^{2}}{3} & =\frac{(a+b)^{2}-a b}{3} \\
& =\frac{1}{3}\left[(a+b)^{2}-\frac{1}{2} a b-\frac{1}{2} a b\right] \\
& \geqslant \sqrt[3]{(a+b)^{2} \cdot\left(-\frac{1}{2} a b\right) \cdot\left(-\frac{1}{2} a b\right)}=\sqrt[3]{\frac{a^{2} b^{2}(a+b)^{2}}{4}},
\end{aligned}
$$

The equality holds if and only if $(a+b)^{2}=-\frac{1}{2} a b$, i.e., $a=-\frac{b}{2}$ or $b=-\frac{a}{2}$.
In summary, $\frac{27}{4} a^{2} b^{2}(a+b)^{2} \leqslant\left(a^{2}+a b+b^{2}\right)^{3}$,

The equality is achieved when $a=b$ and $a=-\frac{b}{2}$ or $b=-\frac{a}{2}$.
Therefore, the required $\lambda=\frac{27}{4}$.