# Inequality of complex numbers in the spirit of Schwarz lemma

I was reading a paper and the autor uses the following lemma of complex analysis

Let $$R_{1},R_{2},\rho \in \mathbb{R}$$ with $$R_{2}>R_{1}>0$$ and $$R_{2}\geq \rho >0$$. Let $$f$$ be a holomorphic function in a open set that contains the closed disc $$\{|z|\leq R_{2}\}$$. If $$f$$ is not identically zero in $$\{|z|\leq R_{2}\}$$, then the number of zeros of $$f$$ in $$\{|z|\leq \rho\}$$, $$n(f,\rho)$$ satisfies $$n(f,\rho) \log \left( \frac{R_{2}^{2}-R_{1}\rho}{R_{2}(R_{1}+\rho)} \right)\leq \log \left( \frac{||f||_{R_{2}}}{||f||_{R_{1}}} \right).$$

The proof is the following

First of all, notice that the set of zeros in $$\{|z|\leq \rho\}$$ can not be infinite, because if this happen, then there will be a accumulation point and then, as $$f$$ is holomorphic, $$f \equiv 0$$. So, we have finite zeros in $$\{|z|\leq \rho\}$$. Let us call them $$z_{1},\ldots,z_{t}$$. The function $$g(z):=f(z) \prod_{j=1}^{t} \frac{R_{2}^{2}-z \overline{z}_{j}}{R_{2}(z-z_{j})}$$ is holomorphic in an open set that contains $$\{|z| \leq R_{2}\}$$. By the maximum modulus principle, $$||g||_{R_{1}}\leq ||g||_{R_{2}}$$. Also, notice that if $$|z|=R_{2}$$ then $$\frac{|R_{2}^{2}-z\overline{z}_{j}|}{|R_{2}(z-z_{j})}| = \frac{|z\overline{z}-z\overline{z}_{j}|}{|R_{2}(z-z_{j})|} =1,$$ so, $$||g||_{R_{2}}=||f||_{R_{2}}$$. Also, using the triangle inequality we obtain that if $$|z_{j}| \leq \rho$$ and $$|z|=R_{1}$$ $$\frac{|R_{2}^{2}-z\overline{z}_{j}|}{|R_{2}(z-z_{j})|} \geq \frac{|R_{2}^{2}-R_{1}\rho|}{|R_{2}(z-z_{j})|} \geq \frac{|R_{2}^{2}-R_{1}\rho|}{|R_{2}(R_{1}+\rho)|}=\frac{R_{2}^{2}-R_{1}\rho}{R_{2}(R_{1}+\rho)},$$ and then, $$||g||_{R_{1}}\geq ||f||_{R_{1}} \left( \frac{R_{2}^{2}-R_{1}\rho}{R_{2}(R_{1}+\rho)} \right)^{t}.$$ Finally, $$||f||_{R_{2}}=||g||_{R_{2}}\geq ||g||_{R_{1}}\geq ||f||_{R_{1}} \left( \frac{R_{2}^{2}-R_{1}\rho}{R_{2}(R_{1}+\rho)} \right)^{t}.$$

Then the author says (without proof) that one can improve the inequality

$$\frac{|R_{2}^{2}-z\overline{z}_{j}|}{|R_{2}(z-z_{j})|} \geq \frac{R_{2}^{2}-R_{1}\rho}{R_{2}(R_{1}+\rho)}$$

to the following:

$$\frac{|R_{2}^{2}-z\overline{z}_{j}|}{|R_{2}(z-z_{j})|} \geq \frac{R_{2}^{2}+R_{1}\rho}{R_{2}(R_{1}+\rho)}.$$

I know this seems trivial, but I can't prove it. I tried the usual tricks of triangle inequality but I only obtained bad results.

By scaling (replacing $$z,z_j$$ by $$\frac{z}{R_2}, \frac{z_j}{R_2}$$) it seems that you want to prove the inequality:

$$z\ne w, |z|=r, |w|=\rho <1$$ implies $$\frac{|1-z\bar w|}{|z-w|}\ge \frac{1+r\rho}{r+\rho}$$.

If we let $$z=re^{i\alpha}, w=\rho e^{i\beta}$$ and we square, this translates to:

$$\frac{1+(r\rho)^2-2r \rho cos(\alpha-\beta)}{r^2+\rho^2-2r \rho cos(\alpha-\beta)}\ge (\frac{1+r\rho}{r+\rho})^2$$

But now we notice the standard inequality $$1+(r\rho)^2 >r^2+\rho^2$$ so renaming the terms by $$b=1+(r\rho)^2, c=r^2+\rho^2,a= 2r \rho>0, b>c, b,c \ge a, \theta =\alpha-\beta$$ we need to find the minimum on $$\theta$$ of :

$$\frac{b-acos(\theta)}{c-a cos(\theta)}=1+\frac{b-c}{c-a cos(\theta)}$$.

But $$c-a cos(\theta) \le c+a$$, so

$$\frac{1+(r\rho)^2-2r \rho cos(\alpha-\beta)}{r^2+\rho^2-2r \rho cos(\alpha-\beta)}=1+\frac{b-c}{c-a cos(\theta)} \ge 1 + \frac{b-c}{c+a}=\frac{b+a}{c+a}=(\frac{1+r\rho}{r+\rho})^2$$

and we are done!