Example 6 Find the largest real number $m$ such that for all positive numbers $a, b, c$ satisfying $a+b+c=1$, we have
$$
10\left(a^{3}+b^{3}+c^{3}\right)-m\left(a^{5}+b^{5}+c^{5}\right) \geqslant 1 \text {. }
$$
(2005, China Girls' Mathematical Olympiad)
【Analysis】Take $a=b=c=\frac{1}{3}$, then
$$
\frac{10}{9}-\frac{m}{3^{4}} \geqslant 1 \Rightarrow m \leqslant 9 \text {. }
$$

Below we prove that $m=9$ satisfies the condition.
Substitution and reduction of power: Let
$$
(u, v)=(a b+b c+c a, a b c) .
$$

Then $0<u \leqslant \frac{1}{3}$, and $a^{2}+b^{2}+c^{2}=1-2 u$.
From the equation
$$
\begin{array}{l}
(x-a)(x-b)(x-c)=0 \\
\Rightarrow x^{3}-x^{2}+u x-v=0 \\
\Rightarrow a^{3}=a^{2}-u a+v .
\end{array}
$$

By cyclic summation, we get
$$
a^{3}+b^{3}+c^{3}=1-3 u+3 v \text {. }
$$

By recursive calculation, we get
$$
\begin{array}{l}
a^{4}+b^{4}+c^{4}=2 u^{2}-4 u+4 v+1, \\
a^{5}+b^{5}+c^{5}=5 u^{2}-5 u v-5 u+5 v+1 .
\end{array}
$$

Thus, inequality (1) is equivalent to
$$
(1-3 u)(u-v) \geqslant 0 \text {. }
$$

Substitution yields
$$
\begin{array}{l}
\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right) \cdot \\
(a b+b c+c a(1-b)) \geqslant 0 .
\end{array}
$$

Therefore, inequality (1) holds.
Hence, the maximum value of the real number $m$ is 9.