Example 2 Let $a$ and $b$ both be positive numbers, and $a+b=1$. Find
$$
E(a, b)=3 \sqrt{1+2 a^{2}}+2 \sqrt{40+9 b^{2}}
$$

the minimum value.
By the Cauchy-Schwarz inequality, we have
$$
\left(2 a^{2}+1\right)\left(\frac{1}{2}+\lambda^{2}\right) \geqslant(a+\lambda)^{2},
$$

where $\lambda$ is a positive constant to be determined.
Equality holds if and only if $4 a^{2}=\frac{1}{\lambda^{2}}$, i.e., $\lambda^{2}=\frac{1}{4 a^{2}}$.
Thus, $\sqrt{1+2 a^{2}} \geqslant \frac{a+\lambda}{\sqrt{\frac{1}{2}+\lambda^{2}}}$.
Again, by the Cauchy-Schwarz inequality, we have
$$
\left(9 b^{2}+40\right)\left(\frac{1}{9}+\frac{1}{40} \mu^{2}\right) \geqslant(b+\mu)^{2},
$$

where $\mu$ is a positive constant to be determined.
Equality holds if and only if $81 b^{2}=\frac{1600}{\mu^{2}}$, i.e., $\mu^{2}=\frac{1600}{81 b^{2}}$.
Thus, $\sqrt{40+9 b^{2}} \geqslant \frac{b+\mu}{\sqrt{\frac{1}{9}+\frac{1}{40} \mu^{2}}}$.
Therefore, $3 \sqrt{1+2 a^{2}}+2 \sqrt{40+9 b^{2}}$
$\geqslant \frac{3(a+\lambda)}{\sqrt{\frac{1}{2}+\lambda^{2}}}+\frac{2(b+\mu)}{\sqrt{\frac{1}{9}+\frac{1}{40} \mu^{2}}}$.
To make the right-hand side of the above inequality a constant, it is necessary and sufficient that
$\frac{3}{\sqrt{\frac{1}{2}+\lambda^{2}}}=\frac{2}{\sqrt{\frac{1}{9}+\frac{1}{40} \mu^{2}}}$.
Given $a>0, b>0, a+b=1$, substituting
$\lambda^{2}=\frac{1}{4 a^{2}}, \mu^{2}=\frac{1600}{81 b^{2}}$,

into equation (1) yields
$$
\begin{array}{l}
\frac{3}{\sqrt{\frac{1}{2}+\frac{1}{4 a^{2}}}}=\frac{2}{\sqrt{\frac{1}{9}+\frac{40}{81 b^{2}}}} \\
\Rightarrow \frac{40}{9 b^{2}}=1+\frac{1}{(1-b)^{2}} \\
\Rightarrow 9 b^{4}-18 b^{3}-22 b^{2}+80 b-40=0 \\
\Rightarrow(b+2)(3 b-2)\left(3 b^{2}-10 b+10\right)=0 \\
\Rightarrow b=\frac{2}{3}, a=\frac{1}{3}, \lambda=\frac{3}{2}, \mu=\frac{20}{3} .
\end{array}
$$

Therefore, when $a=\frac{1}{3}, b=\frac{2}{3}$,
$$
E(a, b)=3 \sqrt{1+2 a^{2}}+2 \sqrt{40+9 b^{2}}
$$

achieves its minimum value $5 \sqrt{11}$.