Let $e(x)=e^{2\pi i x}$ and consider the following functions defined for $x\in [0,1]$: $$ f_1(x)=\frac{e(10x)-e(x)}{e(x)-1}, \quad f_2(x)=\frac{e(110x)-e(11x)}{e(11x)-1}, $$ and $$ f_3(x)=\frac{e(1010x)-e(101x)}{e(101x)-1}\cdot\frac{e(100x)-1}{e(10x)-1} $$ I would like to show bounds of the form $|f_1(x)+f_2(x)| \ll |f_1(x)|$ and $|f_1(x)+f_2(x)+f_3(x)|\ll |f_1(x)+f_2(x)|$.
For the first bound, the zeros of $f_1(x)$ occur at $x=n/9$ for $n\in\{0,\ldots 9\}$ and now note that $f_2(x)$ is also zero at those points. It is then possible to show that $\lim_{x\to n/9} |f_2(x)/f_1(x)|$ exists, so that $|f_2(x)/f_1(x)|$ is continuous and therefore attaches its maximum on $[0,1]$. Hence $$ \frac{|f_1(x)+f_2(x)|}{|f_1(x)|}\le 1+ \frac{|f_2(x)|}{|f_1(x)|} \ll 1. $$ I don't know how to prove the second bound. My intuition tells me that if $f_1(x)+f_2(x)$ is zero, then $f_1(x)+f_2(x)+f_3(x)$ is zero as well but I don't know how to prove this without explicitely computing the zeros. I'm also interested in the implied constant. By plotting the functions on sage, $$ \frac{|f_1(x)+f_2(x)|}{|f_1(x)|}\le 40, $$ and $$ \frac{|f_1(x)+f_2(x)+f_3(x)|}{|f_1(x)+f_2(x)|}\le 60 $$ but I don't know how to prove this without a numeric computation. This functions appear on the context of the Fourier transform of some set of integers