Example 5 Let $a, b, c$ all be positive numbers, $a+b+4c^2=1$, find the maximum value of $\sqrt{a}+\sqrt{b}+\sqrt{2}c$.

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Given that $a, b, c$ are all positive numbers,
we have $a+b+4 c^{2}=(\sqrt{a})^{2}+(\sqrt{b})^{2}+(2 c)^{2}$.
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(a+b+4 c^{2}\right) \cdot\left[1^{2}+1^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\right] \\
=\left[(\sqrt{a})^{2}+(\sqrt{b})^{2}+(2 c)^{2}\right] \cdot\left[1^{2}+1^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\right] \\
\geqslant(\sqrt{a}+\sqrt{b}+\sqrt{2} c)^{2}, \\
\therefore(\sqrt{a}+\sqrt{b}+\sqrt{2} c)^{2} \leqslant 1 \times \frac{5}{2}=\frac{5}{2}, \\
\text { i.e., } \sqrt{a}+\sqrt{b}+\sqrt{2} c \leqslant \frac{\sqrt{10}}{2} .
\end{array}$$

Equality holds $\Leftrightarrow \frac{\sqrt{a}}{1}=\frac{\sqrt{b}}{1}=\frac{2 c}{\frac{1}{\sqrt{2}}}$.
Solving the system of equations $\left\{\begin{array}{l}\sqrt{a}=\sqrt{b}=2 \sqrt{2} c, \\ a+b+4 c^{2}=1,\end{array}\right.$ we get $\left\{\begin{array}{l}a=\frac{2}{5}, \\ b=\frac{2}{5}, \\ c=\frac{\sqrt{5}}{10} .\end{array}\right.$
Therefore, when $a=b=\frac{2}{5}, c=\frac{\sqrt{5}}{10}$, $\sqrt{a}+\sqrt{b}+\sqrt{2} c$ attains its maximum value
$$\frac{\sqrt{10}}{2}$$