Note that $\frac{1}{16} f(2 x)=x^{4}+7 x^{3}+13 x^{2}+7 x+1$. 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 $\{1,2,3,4\}$ : $|z_{\sigma(1)} z_{\sigma(2)}+z_{\sigma(3)} z_{\sigma(4)}| \leq \frac{1}{2}(z_{\sigma(1)} z_{\sigma(2)}+z_{\sigma(3)} z_{\sigma(4)}+z_{\sigma(4)} z_{\sigma(3)}+z_{\sigma(2)} z_{\sigma(1)})$ Let the roots be ordered $z_{1} \leq z_{2} \leq z_{3} \leq z_{4}$, 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 8.