Proof
$$(17) \Leftrightarrow 4 b^{3} c^{3} \geqslant(b+c)^{2}\left[c^{2}-(a-b)^{2}\right]\left[b^{2}-(c-a)^{2}\right] \Leftrightarrow$$
$$4 b^{3} c^{3} \geqslant b^{2} c^{2}(b+c)^{2}+(b+c)^{2}(a-b)^{2}(c-a)^{2}-$$
$$b^{2}(b+c)^{2}(b-a)^{2}-c^{2}(b+c)^{2}(c-a)^{2} \Leftrightarrow$$
$$b^{2}(b+c)^{2}(b-a)^{2}+c^{2}(b+c)^{2}(c-a)^{2} \geqslant$$
$$b^{2} c^{2}(b-c)^{2}+(b+c)^{2}(a-b)^{2}(c-a)^{2} \Leftrightarrow$$
$$\left[b^{3}(b+c)(b-a)^{2}+b^{3} c(b-a)^{2}+b^{2} c^{2}(b-a)^{2}\right]+$$
$$\left[c^{3}(b+c)(c-a)^{2}+b c^{3}(c-a)^{2}+b^{2} c^{2}(c-a)^{2}\right] \geqslant$$
$$\left[b^{2} c^{2}(b-a)^{2}+b^{2} c^{2}(a-c)^{2}+2 b^{2} c^{2}(b-a)(a-c)\right]+$$
$$(b+c)^{2}(b-a)^{2}(c-a)^{2} \Leftrightarrow$$
$$\left[b^{3}(b+c)(b-a)^{2}+c^{3}(b+c)(c-a)^{2}-(b+c)^{2}(b-a)^{2}(c-a)^{2}\right]+$$
$$b c\left[b^{2}(b-a)^{2}+c^{2}(c-a)^{2}+2 b c(b-a)(c-a)\right] \geqslant 0 \Leftrightarrow$$
$$(b+c)\left[b(b+c-a)(b-c+a)(b-a)^{2}+\right.$$
$$\left.c(c+b-a)(c-b+a)(c-a)^{2}\right]+$$
$$b c[b(b-a)+c(c-a)]^{2} \geqslant 0$$
This inequality is obviously true.
Note 1. Equation (18) is equivalent to
$$2 R w_{a} \geqslant a(-a+b+c)$$
where $R$ is the circumradius of $\triangle ABC$, and $w_{a}$ is the angle bisector of side $a$.