Example 10 If $a+b+c=1, a, b, c \in \mathbf{R}^{+}$, $\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1} \geqslant m$, find the maximum value of $m$.

 untranslated text remains unchanged.
Let $f(a, b, c) = \sqrt{4a + 1} + \sqrt{4b + 1} + \sqrt{4c + 1}$, and assume $a, b, c \geqslant 0$.

By symmetry, without loss of generality, assume $a \geqslant b \geqslant c$, and let $a' = a + c$, $c' = 0$,

then $f(a, b, c) - f(a', b, c') = \sqrt{4a + 1} + \sqrt{4c + 1} - \sqrt{4(a + c) + 1} - 1$,

and $(\sqrt{4a + 1} + \sqrt{4c + 1})^2 - (\sqrt{4(a + c) + 1} + 1)^2 = 2\sqrt{(4a + 1)(4c + 1)} - 2\sqrt{4(a + c) + 1} \geqslant 0$,

when $a + b = 1$,
\[
\begin{array}{l}
(\sqrt{4a + 1} + \sqrt{4b + 1})^2 \\
= 4(a + b) + 2 + 2\sqrt{(4a + 1)(4b + 1)} \\
= 6 + 2\sqrt{16ab + 4(a + b) + 1} \\
= 6 + 2\sqrt{16ab + 5} \\
\geqslant 6 + 2\sqrt{5} \\
= (1 + \sqrt{5})^2,
\end{array}
\]

the equality holds when one of $a, b$ is zero,
thus, $\sqrt{4a + 1} + \sqrt{4b + 1} + \sqrt{4c + 1} \geqslant 2 + \sqrt{5}$, the equality holds when two of $a, b, c$ are zero.

When $a \rightarrow 0, b \rightarrow 0, c \rightarrow 1$, $\sqrt{4a + 1} + \sqrt{4b + 1} + \sqrt{4c + 1} \rightarrow 2 + \sqrt{5}$, so $m_{\max} = 2 + \sqrt{5}$.