Bound on the sum of arguments 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 the 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? 
I think tools provided in arXiv:0907.2960 [math.GN] may be useful here. 

Fedor Petrov: your first inequality, 
$$\operatorname{arg}\frac{1−zf(s−u)}{1−zf(s+u)}<π/2−\operatorname{arg}(1+z\bar{z}f(2u))$$
(as well as the second one, of course) is incorrect, in general. E.g., take $s=0$, $u=\pi/2-h$, $h\downarrow0$, $0<z<1$, $1-z=o(h)$, and $f(t)=e^{it}$ for all real $t$. Then the left-hand side of your inequality goes to $\pi/2$, whereas its right-hand side goes to $0$. The identity 
$$
2\Re (1-A)(1-\bar{B})(1+C)=|-C+\bar{A}+B-1|^2+\det M>0
$$
in your last display is incorrect as well, and even the inequality you apparently wanted to get from that identity is also in general incorrect. E.g., for $A=-i$, $B=1+i$, and $C=i$, one has $2\Re (1-A)(1-\bar{B})(1+C)=-4<-1=\det M$. What I found somewhat fascinating here, though, is that the above identity would be true if one could change just the sign of one monomial in the polynomial expansion (in the real and and imaginary parts of $A,B,C$) of the difference between the left-hand side and the right-hand side of this incorrect identity. 
Of course, one can rewrite the nonstrict version of the inequality (*) that I proposed as $\Psi_{f,z}(u)+\Psi_{f,z}(-u)\le\pi$, where 
$$\Psi_{f,z}(u):=\sup_{s\in\mathbb{R}}\arg\frac{1-zf(s-u)}{1-zf(s+u)}, 
$$
and then try to find a manageable upper bound $\Theta_{f,z}(u)$ on $\Psi_{f,z}(u)$ such that $\Theta_{f,z}(u)+\Theta_{f,z}(-u)\le\pi$. However, I would be greatly surprised if such a simple bound as your $\pi/2- \arg(1+z\bar{z}f(2u))$ would do. 
I still think the best bet so far is to try to use tools provided in arXiv:0907.2960 [math.GN]. 
As for your question about what real algebraic geometry has to do with this problem: real algebraic geometry is to a large extent about solving systems of polynomial inequalities over $\mathbb{R}$ (including various Positivstellensatzen), which is clearly relevant here. 
 A: Let's prove that
$$
\arg \frac{1-zf(s-u)}{1-zf(s+u)}< \pi/2- \arg(1-z\bar{z}f(2u)).
$$
Then summing this up with an analogous inequality 
$$
\arg \frac{1-zf(t+u)}{1-zf(t-u)}< \pi/2- \arg(1-z\bar{z}f(-2u))
$$
we get what we need. 
Denote $zf(s-u)=A$, $zf(s+u)=\bar{B}$, $z\bar{z}f(2u)=C$. Then for functions $\varphi_1(x)=e^{isx}$, $\varphi_2(x)=\bar{z}e^{iux}$, $\varphi_3(x)=\bar{z}e^{-iux}$ in $(L^2,\mu)$ we have $A=\langle\varphi_1,\varphi_2\rangle$, $\bar{B}=\langle\varphi_1,\varphi_3\rangle$, $C=\langle\varphi_2,\varphi_3\rangle$. Thus the following matrix is non-negative definite as the Gram matrix of our functions
$$
\pmatrix{\|\varphi_1\|^2&A&\bar{B}\\\bar{A}&\|\varphi_2\|^2&C\\B&\bar{C}&\|\varphi_3\|^2}.
$$
We may increase diagonal elements upto 1. Of course, matrix remains non-negative definite and, moreover, becomes positive definite (since $|z|<1$, we strictly increase two diagonal elements of our matrix, and its determinant becomes strictly positive):
$$
M=\pmatrix{1&A&\bar{B}\\\bar{A}&1&C\\B&\bar{C}&1},\,\det M>0.
$$
What we have to prove is that $\arg (1-A)(1-B)=\arg \frac{1-A}{1-\bar{B}}<\pi/2-\arg(1-C)$. Assume the contrary, i.e. $\arg(1-A)+\arg(1-B)\geq \pi/2-\arg(1-C)$. Since $\arg(1-A),\arg(1-B),\arg(1-C)\in (-\pi/2,\pi/2)$ we get $\theta:=\arg (1-A)+\arg (1-B)+\arg (1-C)\in [\pi/2,3\pi/2]$, i.e. $(1-A)(1-B)(1-C)=e^{i\theta}r$, $r>0$, $\Re (1-A)(1-B)(1-C)=r\cos \theta\leq 0$. 
After this point there should be more elegant way to finish the proof, but let me provide at least some way. 
First of all, 
$$
\det M=1-|A|^2-|B|^2-|C|^2+2\Re(ABC)=(1-|A|^2)(1-|B|^2)-|C-\bar{A}\cdot \bar{B}|^2> 0.
$$
Thus $C=\bar{A}\cdot \bar{B}+w$, $|w|< R:=\sqrt{(1-|A|^2)(1-|B|^2)}$. Under these conditions the best lower estimate for $\Re (1-A)(1-B)(1-C)$ is 
$$
\Re (1-A)(1-B)(1-C)>\Re (1-A)(1-B)(1-\bar{A}\cdot \bar{B})-R\cdot |1-A|\cdot |1-B|.
$$
Now we have
\begin{align*}
X:=\Re (1-A)(1-B)(1-\bar{A}\cdot \bar{B})=\Re (1-A)(1-B)\left((1-\bar{A})+\bar{A}\cdot (1-\bar{B})\right)=\\
=|1-A|^2(1-\Re B)+|1-B|^2(\Re A-|A|^2)
\end{align*}
Analogously $X=|1-A|^2(\Re B-|B|^2)+|1-B|^2(1-\Re A)$. Taking half sum of two expressions for $X$ and applying AM-GM for two summands we get
$$
X=\frac{|1-A|^2(1-|B|^2)+|1-B|^2(1-|A|^2)}2\geq \sqrt{(1-|B|^2)(1-|A|^2)}\cdot |1-A|\cdot |1-B|,
$$
hence $\Re (1-A)(1-B)(1-C)>0$, a contradiction.
