A difficult inequality involving complex numbers

The easiest (in the sense of "least work") way to prove it uses a tiny bit of complex analysis.

Consider the function

$$f(z) = \sum_{i=1}^n \frac{1}{1 - z_i\cdot z}.$$

This is a meromorphic function having simple poles at $\frac{1}{z_i},\, 1 \leqslant i \leqslant n$, and nowhere else. The Taylor series of $f$ with centre $0$,

$$f(z) = \sum_{i=1}^n \sum_{r=0}^\infty z_i^r\cdot z^r = \sum_{r=0}^\infty \left(\sum_{i=1}^n z_i^r\right)z^r,$$

therefore has radius of convergence

$$R = \min_{1\leqslant i\leqslant n} \left\lvert \frac{1}{z_i}\right\rvert = \frac{1}{\max\limits_{1\leqslant i \leqslant n} \lvert z_i\rvert}.$$

But the assumption is that the coefficients of the Taylor series are bounded, hence the radius of convergence is at least $1$.

This is a partial answer. Suppose without loss of generality that $z_1$ has maximum absolute value among $z_1, \dots, z_n$. If we assume that $|z_1| > |z_i|$ for all $i=2, \dots, n$ it's easy to see that $|z_1| \leq 1$.

Otherwise, suppose by contradiction $|z_1| > 1$.

Now, for all $r \geq 1$ the following inequality holds $$C \geq |z_1^r + \dots + z_n^r| \geq |z_1|^r - |z_2|^r - \dots - |z_n|^r$$

dividing by $|z_1|^r$ we get

$$\frac{C}{|z_1|^r} \geq 1 -\frac{|z_2|^r}{|z_1|^r} - \dots \frac{|z_n|^r}{|z_1|^r}$$

sending $r \to + \infty$ we get the contradiction $0 \geq 1 - 0- \dots -0=1$.

Maybe the general case can be treated by using this particular case.