14. Find the smallest real number $m$, such that for any positive real numbers $a, b, c$ 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)
14. When $a=b=c=\frac{1}{3}$, we have $m \geqslant 27$.

Below, we prove the inequality $27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1$ for any positive real numbers $a, b, c$ satisfying $a+b+c=1$.

Since $a, b, c$ are positive real numbers, we have $a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, c^{3}+a^{3} \geqslant c^{2} a+c a^{2}$. Adding these three inequalities, we get
$$\begin{array}{c}
2\left(a^{3}+b^{3}+c^{3}\right) \geqslant a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b \\
3\left(a^{3}+b^{3}+c^{3}\right) \geqslant a^{3}+b^{3}+c^{3}+a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b= \\
\left(-a^{2}+b^{2}+c^{2}\right)(a+b+c)=a^{2}+b^{2}+c^{2}
\end{array}$$

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

Therefore, the minimum value of $m$ is 27.