Prove that by Hölder's inequality,
$$\left(\sum_{\text {cyc }} \sqrt{\frac{a(b+c)}{(2 b+c)(b+2 c)}}\right)^{2} \sum_{\text {cyc }} \frac{a^{2}(2 b+c)(b+2 c)}{b+c} \geqslant(a+b+c)^{3},$$

Therefore, it suffices to prove that
$$\begin{array}{l} (a+b+c)^{3} \geqslant 2 \sum_{\mathrm{cyc}} \frac{a^{2}(2 b+c)(b+2 c)}{b+c} . \\ \text { In fact, }(a+b+c)^{3}-2 \sum_{\mathrm{cyc}} \frac{a^{2}(2 b+c)(b+2 c)}{2 b+c}=\sum_{\mathrm{cyc}} \\ \frac{(b-c)^{2}(b+c-a)^{2}}{2(b+c)} \geqslant 0 . \end{array}$$

In conclusion, the original inequality holds.