8. Find the smallest real number $m$, such that for any positive integers $a, b, c$ satisfying $a+b+c=1$, the inequality $m\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1$ holds.
(2006 China Southeast Mathematical Olympiad)
8. First find the minimum value of $m$, then provide a proof.

When $a=b=c=\frac{1}{3}$, from the given inequality we get $m \geqslant 27$.
Next, we prove: 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
$$
a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, \quad b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, \quad c^{3}+a^{3} \geqslant c^{2} a+c a^{2}.
$$

Adding these three inequalities, we get
$$
\begin{array}{l}
2\left(a^{3}+b^{3}+c^{3}\right) \geqslant a^{2} b+b^{2} c+c^{2} a+a b^{2}+b c^{2}+c a^{2}. \\
\text{Then } 3\left(a^{3}+b^{3}+c^{3}\right) \geqslant a^{3}+b^{3}+c^{3}+a^{2} b+b^{2} c+c^{2} a+a b^{2}+b c^{2}+c a^{2} \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)=a^{2}+b^{2}+c^{2}.
\end{array}
$$
$$
\begin{array}{l}
\text{Thus } 6\left(a^{2}+b^{2}+c^{2}\right)+1=6\left(a^{2}+b^{2}+c^{2}\right)+(a+b+c)^{2} \\
\leqslant 6\left(a^{2}+b^{2}+c^{2}\right)+3\left(a^{2}+b^{2}+c^{2}\right) \\
=9\left(a^{2}+b^{2}+c^{2}\right) \leqslant 27\left(a^{3}+b^{3}+c^{3}\right).
\end{array}
$$