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.