Prove that the original inequality is equivalent to
$$\begin{array}{l} \sum \frac{a}{b+c}\left[a^{2}+3 b c-(b+c)^{2}\right] \geqslant 0 \Leftrightarrow \\ \frac{a\left(a^{2}+3 b c\right)}{b+c}+\frac{b\left(b^{2}+3 c a\right)}{c+a}+\frac{c\left(c^{2}+3 a b\right)}{a+b} \geqslant 2(a b+b c+c a) \end{array}$$

Using the Cauchy-Schwarz inequality,
$$\begin{aligned} \text { Left side }= & \frac{a^{2}\left(a^{2}+3 b c\right)}{a(b+c)}+\frac{b^{2}\left(b^{2}+3 c a\right)}{b(c+a)}+\frac{c^{2}\left(c^{2}+3 a b\right)}{c(a+b)} \geqslant \\ & \frac{\left(a \sqrt{a^{2}+3 b c}+b \sqrt{b^{2}+3 c a}+c \sqrt{c^{2}+3 a b}\right)^{2}}{2(a b+b c+c a)} \end{aligned}$$

Thus, it suffices to prove
$$a \sqrt{a^{2}+3 b c}+b \sqrt{b^{2}+3 c a}+c \sqrt{c^{2}+3 a b} \geqslant 2(a b+b c+c a)$$

Using the AM-GM inequality, we have
$$\sum a \sqrt{a^{2}+3 b c}=\sum \frac{a(b+c)\left(a^{2}+3 b c\right)}{(b+c) \sqrt{a^{2}+3 b c}} \geqslant 2 \sum \frac{a(b+c)\left(a^{2}+3 b c\right)}{a^{2}+3 b c+(b+c)^{2}}$$

It remains to prove
$$\begin{array}{l} 2 \sum \frac{a(b+c)\left(a^{2}+3 b c\right)}{a^{2}+3 b c+(b+c)^{2}} \geqslant 2 \sum a b \Leftrightarrow \\ 2 \sum \frac{a(b+c)\left(a^{2}+3 b c\right)}{a^{2}+3 b c+(b+c)^{2}} \geqslant \sum a(b+c) \end{array}$$

Let $p = a^{2} + b^{2} + c^{2}$, then $\Leftrightarrow$
$$\begin{array}{l} 2 \sum \frac{a(b+c)\left(a^{2}-b^{2}-c^{2}+b c\right)}{p+5 b c} \geqslant 0 \Leftrightarrow \\ \sum \frac{a^{3}(b+c)-a\left(b^{3}+c^{3}\right)}{p+5 b c} \geqslant 0 \Leftrightarrow \\ \sum \frac{a b\left(a^{2}-b^{2}\right)}{p+5 b c}-\sum \frac{c a\left(c^{2}-a^{2}\right)}{p+5 b c} \geqslant 0 \Leftrightarrow \\ \sum \frac{a b\left(a^{2}-b^{2}\right)}{p+5 b c}-\sum \frac{a b\left(a^{2}-b^{2}\right)}{p+5 c a} \geqslant 0 \Leftrightarrow \\ 5 a b c \sum \frac{(a-b)^{2}(a+b)}{(p+5 b c)(p+5 c a)} \geqslant 0 \end{array}$$

This is clearly true, with equality holding if and only if $a = b = c$ or $a = b, c = 0$ cyclically.