Problem 1612

Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, then the original inequality becomes
$$\sum \frac{x^{2}}{y+z} \geqslant \frac{3}{2}+\frac{3}{4} \sum \frac{(y-z)^{2}}{y+z}$$

Let $p=x+y+z, q=xy+yz+zx, r=xyz=1$, then the above inequality is equivalent to
$$4 p^{4}-18 p^{2} q+12 q^{2}+34 p r-6 p q+6 r \geqslant 0$$

Let $f(q)=4 p^{4}-18 p^{2} q+12 q^{2}+34 p r-6 p q+6 r$, then $f^{\prime}(q)=-6\left(3 p^{2}+p-4 q\right) \leqslant 0$, so by Corollary 10.15(1) (here $n=3, m=0, p=2$), the minimum value of $f(q)$ is achieved when $x=y \leqslant z$, then $x \leqslant 1$, substituting into the inequality we need to prove
$$\frac{2 x^{2}}{x+z}+\frac{z^{2}}{2 x} \geqslant \frac{3}{2}+\frac{3}{2} \cdot \frac{(z-x)^{2}}{z+x}$$

At this point, we also have $x^{2} z=1$, substituting into the above inequality and expanding, we need to prove
$$\begin{array}{l} x^{9}-3 x^{8}+6 x^{6}-3 x^{5}-2 x^{3}+1 \geqslant 0 \Leftrightarrow \\ (x-1)^{2}\left(x^{7}-x^{6}-3 x^{5}+x^{4}+2 x^{3}+3 x^{2}+2 x+1\right) \geqslant 0 \end{array}$$

Since $0 \leqslant x \leqslant 1$, the above inequality clearly holds.
Therefore, the original inequality holds, with equality if and only if $a=b=c=1$.