1.5.13 $\star \star$ Find the largest real number $k$, such that for any positive real numbers $a, b, c$, we have
$$
\begin{aligned}
& \frac{(b-c)^{2}(b+c)}{a}+\frac{(c-a)^{2}(c+a)}{b}+\frac{(a-b)^{2}(a+b)}{c} \\
\geqslant & k\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)
\end{aligned}
$$
Parse: In (1), let $a=b=1$, we get $2(1-c)^{2}(1+c) \geqslant k\left(1-c^{2}\right)$. Let $c \rightarrow 0^{+}$, we can get $k \leqslant 2$. Below, we prove that when $k=2$, inequality (1) holds, that is, to prove
$$
\sum \frac{(b-c)^{2}(b+c)}{a} \geqslant 2\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=\sum(b-c)^{2} .
$$

By symmetry, without loss of generality, assume $a \geqslant b \geqslant c$, note that,
$$
\text { (2) } \Leftrightarrow \sum \frac{(b-c)^{2}(b+c-a)}{a} \geqslant 0 \text {. }
$$

If $b+c-a \geqslant 0$, then each term on the left side of (3) is not less than zero, the inequality obviously holds. If $b+c-a<0$, then
$$
\frac{(b-c)^{2}(b+c-a)}{a}+\frac{(a-c)^{2}(a+c-b)}{b} \geqslant \frac{(b-c)^{2}(b+c-a)}{a}+
$$
$\frac{(b-c)^{2}(a+c-b)}{a}=\frac{2 c(b-c)^{2}}{a} \geqslant 0$, and $\frac{(a-b)^{2}(a+b-c)}{c} \geqslant 0$. Therefore, inequality (3) still holds.
In summary, the maximum real number $k=2$.