Proof:
$$\begin{aligned} & \sum_{\text {cyc }}\left(a^{3}+a b c\right) \geqslant \sum_{\text {cyc }} a b \sqrt{2\left(a^{2}+b^{2}\right)} \\ \Leftrightarrow & \sum_{\text {cyc }}\left(a^{6}+2 a^{8} b^{3}+6 a^{4} b c+3 a^{2} b^{2} c^{2}-2 a^{4} b^{2}-2 a^{4} c^{2}\right) \\ \geqslant & 4 \sum_{\text {cyc }} a^{2} b c \sqrt{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)} \end{aligned}$$

Since
$$2 \sqrt{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)} \leqslant 2 a^{2}+b^{2}+c^{2},$$

to prove the original inequality, it suffices to prove
$$\begin{aligned} & \sum_{\text {cyc }}\left(a^{6}-2 a^{4} b^{2}-2 a^{4} c^{2}+2 a^{3} b^{3}\right. \\ + & \left.2 a^{4} b c-2 a^{3} b^{2} c-2 a^{3} c^{2} b+3 a^{2} b^{2} c^{2}\right) \geqslant 0 . \end{aligned}$$

Using the Sum of Squares (S. O. S.) method, it is easy to prove equation (14):
$$\begin{aligned} & \sum_{\text {cyc }}\left(a^{6}-2 a^{4} b^{2}-2 a^{4} c^{2}+2 a^{3} b^{3}+2 a^{4} b c-2 a^{3} b^{2} c-2 a^{3} c^{2} b\right. \\ & \left.+3 a^{2} b^{2} c^{2}\right) \geqslant 0 \\ \Leftrightarrow & \sum_{\text {cyc }}\left(2 a^{6}-4 a^{4} b^{2}-4 a^{4} c^{2}+4 a^{3} b^{3}+4 a^{4} b c-4 a^{3} b^{2} c-4 a^{3} c^{2} b\right. \\ & \left.+6 a^{2} b^{2} c^{2}\right) \geqslant 0 \\ \Leftrightarrow & \sum_{\text {cyc }}\left(a^{6}-a^{4} b^{2}-a^{2} b^{4}+b^{6}\right)-3 \sum_{\text {cyc }}\left(a^{4} b^{2}-2 a^{3} b^{3}+a^{2} b^{4}\right) \\ & -\sum_{\text {cyc }}\left(c^{3} a^{3}-c^{3} a^{2} b-c^{3} a b^{2}+c^{3} b^{3}\right)+2 \sum_{\text {cyc }}\left(a^{4} b c-a^{3} b^{2} c\right. \\ & \left.-b^{3} a^{2} c+b^{4} a c\right)-3 \sum_{\text {cyc }}\left(a^{3} c^{2} b-2 a^{2} b^{2} c^{2}+b^{3} c^{2} a\right) \geqslant 0 \\ \Leftrightarrow & \sum_{\text {cyc }}(a-b)^{2}\left((a+b)^{2}\left(a^{2}+b^{2}\right)-3 a^{2} b^{2}-c^{3}(a+b)\right. \\ & \left.+2 a b c(a+b)-3 a b c^{2}\right) \geqslant 0 \end{aligned}$$

Assume without loss of generality that $a \geqslant b \geqslant c$, then $S_{c} \geqslant 0$,
$$\begin{array}{l} S_{b}=(a+c)^{2}\left(a^{2}+c^{2}\right)-3 a^{2} c^{2}-b^{3}(a+c)+2 a b c(a+c)-3 a b^{2} c \geqslant 0, \\ S_{b}+S_{a} \geqslant(a-b)^{2}\left(a^{2}+b^{2}-c^{2}+a b+a c+b c\right) \geqslant 0 \end{array}$$

Therefore, the original inequality is
$$\sum_{c y c}(a-b)^{2} S_{c} \geqslant S_{a}(b-c)^{2}+S_{b}(b-c)^{2} \geqslant 0$$

In conclusion, the original inequality holds.