Example 5 Find the smallest real number $m$, such that for any positive numbers $a, b, c$ satisfying $a+b+c=1$, we have
$$
m\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1 \text {. }
$$
(3rd China Southeast Mathematical Olympiad)
Explanation: First find the minimum value of $m$, then provide a proof.
When $a=b=c=\frac{1}{2}$, from the given inequality we have
$$
m \geqslant 27.
$$

Next, we prove that the inequality
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1,
$$

holds for any positive real numbers $a, b, c$ satisfying $a+b+c=1$.
It is easy to see that for any positive real numbers $a, b, c$, we have
$$
\begin{array}{l}
a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, \\
b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, \\
c^{3}+a^{3} \geqslant c^{2} a+c a^{2}.
\end{array}
$$

Adding these three inequalities, we get
$$
\begin{array}{l}
2\left(a^{3}+b^{3}+c^{3}\right) \\
\geqslant a^{2} b+b^{2} c+c^{2} a+a b^{2}+b c^{2}+c a^{2}.
\end{array}
$$

Thus,
$$
3\left(a^{3}+b^{3}+c^{3}\right)
\begin{aligned}
\geqslant & a^{3}+b^{3}+c^{3}+a^{2} b+b^{2} c+c^{2} a+ \\
& a b^{2}+b c^{2}+c a^{2} \\
= & (a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \\
= & a^{2}+b^{2}+c^{2}.
\end{aligned}
$$

Therefore,
$$
6\left(a^{2}+b^{2}+c^{2}\right)+1
\begin{array}{l}
=6\left(a^{2}+b^{2}+c^{2}\right)+(a+b+c)^{2} \\
\leqslant 6\left(a^{2}+b^{2}+c^{2}\right)+3\left(a^{2}+b^{2}+c^{2}\right) \\
=9\left(a^{2}+b^{2}+c^{2}\right) \\
\leqslant 27\left(a^{3}+b^{3}+c^{3}\right).
\end{array}
$$

Thus, the minimum value of $m$ is 27.
Note: First, by setting $a=b=c=\frac{1}{3}$, we get $m \geqslant 27$, then we prove that 27 is the minimum value of $m$. The proof starts from the inequalities $a^{3}+b^{3} \geqslant a^{2} b+a b^{2}$, etc., to derive
$$
3\left(a^{3}+b^{3}+c^{3}\right) \geqslant a^{2}+b^{2}+c^{2},
$$

and thus prove the conclusion.