(1) Find the smallest real number $m$, such that for any positive real numbers $a, b, c$ satisfying $a+b+c=1$, we have
$m\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1$. (Xiong Bin)
Method 1 When $a=b=c=\frac{1}{3}$, we have $m \geqslant 27$.
Next, we prove the inequality
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1
$$

for any positive real numbers $a, b, c$ satisfying $a+b+c=1$.
Since for $0<x<1$, we have
$$
\begin{aligned}
27 x^{3} & \geqslant 6 x^{2}+5 x-\frac{4}{3} \\
& \Leftrightarrow 81 x^{3}-18 x^{2}-15 x+4 \geqslant 0 \\
& \Leftrightarrow(3 x-1)^{2}(9 x+4) \geqslant 0,
\end{aligned}
$$

thus
$$
27 x^{3} \geqslant 6 x^{2}+5 x-\frac{4}{3}, \quad 0<x<1
$$

Therefore,
$$
\begin{array}{l}
27 a^{3} \geqslant 6 a^{2}+5 a-\frac{4}{3}, \\
27 b^{3} \geqslant 6 b^{2}+5 b-\frac{4}{3}, \\
27 c^{3} \geqslant 6 c^{2}+5 c-\frac{4}{3},
\end{array}
$$

Adding the above three inequalities, we get
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1.
$$

Therefore, the minimum value of $m$ is 27.
Method 2 When $a=b=c=\frac{1}{3}$, we have $m \geqslant 27$.
Next, we prove the inequality
$$
27\left(a^{3}+b^{3}+c^{3}\right) \geqslant 6\left(a^{2}+b^{2}+c^{2}\right)+1
$$

for any positive real numbers $a, b, c$ satisfying $a+b+c=1$.
Since $(a-b)^{2}(a+b) \geqslant 0$, we have $a^{3}+b^{3} \geqslant a^{2} b+a b^{2}$, similarly,
$$
b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, \quad c^{3}+a^{3} \geqslant c^{2} a+c a^{2},
$$

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

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.