Maximize $$f(x,y,z)=\frac{1}{|z|\sqrt x}\left|a\left(e^{iz \left(n-\frac12\right)}-e^{iz \left(n-\frac12-y\right)}\right)+b \left(e^{iz \left(n+\frac12\right)}-e^{iz \left(n-\frac12\right)}\right)+c\left(e^{iz \left(n-\frac12+x-y\right)}-e^{iz \left(n+\frac12\right)}\right)\right|$$ subject to $$\mathcal T=\left\{(x,y,z)\, |\, 0\leq y\leq x-1\, , \, x\in[1,2]\, , \, z\in\mathbb R \right\}$$ where $a,b,c\in\mathbb R$, $n\in\mathbb N$.
In the following I will try to develop at least a partial solution. We can rewrite $f$ as $$f(x,y,z)=\frac{1}{\sqrt x}\left |a \int_{n-\frac12-y}^{n-\frac12} e^{iz t} dt +b\int_{n-\frac12}^{n+\frac12} e^{iz t} dt+c\int_{n+\frac12}^{n-\frac12+x-y} e^{iz t} dt\right|$$ If $a,b,c>0$, after applying the triangle inequality (which is exact when $z=0$ because of the sign of a, b and c) we obtain $$f(x,y,z)\leq \frac{1}{\sqrt x}\left(a y+ b+c(x-y-1) \right)=g(x,y)$$ Now it is a lot easier to maximize $f$ because we have to find the maximum of the function $(x,y)\to g(x,y)$ in the triangle: $\left\{(x,y)\, |\, 0\leq y\leq x-1\, , \, x\in[1,2]\right\}$.
However, I can't figure out how manage the general case $a,b,c\in\mathbb R$. Any suggestions?