Prove that for $|z| < R$, we have $\frac{|z|}{|z^2-n^2|} \leq \frac{R}{n^2-R^2}$

Question

Prove that for $|z| < R$, we have $\frac{|z|}{|z^2-n^2|} \leq \frac{R}{n^2-R^2}$, where $n$ is an integer greater than $R$.

This is from page 168, Lang's Complex Analysis. I am having trouble verifying this. Certainly, we have that $|n-z| \geq n - |z| \geq n-R$, but since $|z| \geq |z+n|-n \not\geq R$ I am not sure what to do.

Answer 1

Hint:

\begin{align} |z^2-n^2|\geq \left|n^2-|z|^2\right| \geq n^2-R^2 \end{align}