Problem: Show that for all real $s,t,u$ and all complex $z$ with $|z|<1$ one has $$(*)\qquad \arg\frac{1-zf(s-u)}{1-zf(s+u)} +\arg\frac{1-zf(t+u)}{1-zf(t-u)}<\pi, $$
where $f$ is a characteristic function of a probability distribution $\mu$, so that $f(t)=\int_{-\infty}^\infty e^{itx}\mu(dx)$ for all real $t$, and, as usual, $\arg w\in(-\pi,\pi]$ for any nonzero complex number $w$.

Comments: \ Letting $(t_1,\dots,t_4):=(s,t,u,-u)$, note that the matrix $M:=(f(t_k-t_j))_{k,j=1}^4$ is Hermitian and nonnegative definite or, equivalently, its principal minors (or just the leading principal minors) are nonnegative. This condition on the minors can be rewritten as a system of polynomial inequalities in the real variables $a_{k,j}:=\Re f(t_k-t_j)$ and $b_{k,j}:=\Im f(t_k-t_j)$, with $k,j=1,\dots,4$ and $k>j$. Note that $f(s-u),f(s+u),f(t+u),f(t-u)$ are elements of the matrix $M$.

Note also that inequality (*) can be expressed as a logical formula whose terms are polynomial inequalities in the real and imaginary parts of the complex numbers $z,f(s-u),f(s+u),f(t+u),f(t-u)$.

Numerical experiments suggest that the mentioned condition on the minors together with the condition $|z|<1$ are enough for inequality (*), on the sum of the arguments, to hold. In principle, this can of course be verified purely algorithmically, but it appears to take just too much computing resources for the calculation to complete. Any other idea?