jmc

Note that $$\frac{1}{16}f(2x)=x^4+7x^3+13x^2+7x+1.$$Because the coefficients of this polynomial are symmetric, if $r$ is a root of $f(x)$ then $\frac{4}{r}$ is as well. Further, $f(-1)=-1$ and $f(-2)=16$ so $f(x)$ has two distinct roots on $(-2,0)$ and two more roots on $(-\infty ,-2)$. Now, if $\sigma$ is a permutation of $\{1,2,3,4\}$: $$|z_{\sigma (1)}{z}_{\sigma (2)}+z_{\sigma (3)}{z}_{\sigma (4)}|\le \frac{1}{2}(z_{\sigma (1)}{z}_{\sigma (2)}+z_{\sigma (3)}{z}_{\sigma (4)}+z_{\sigma (4)}{z}_{\sigma (3)}+z_{\sigma (2)}{z}_{\sigma (1)}).$$Let the roots be ordered $z_1\le z_2\le z_3\le z_4$, then by rearrangement the last expression is at least: $$\frac{1}{2}(z_1{z}_4+z_2{z}_3+z_3{z}_2+z_4{z}_1).$$Since the roots come in pairs $z_1{z}_4=z_2{z}_3=4$, our expression is minimized when $\sigma (1)=1,\sigma (2)=4,\sigma (3)=3,\sigma (4)=2$ and its minimum value is $\boxed{8}$.