Show that if
$$P(z) = {a_n}z^{n} + {a_{n-1}}z^{n-1} + ... + {a_1}z + {a_0},$$
is a polynomial of degree $n$ then there is ${R_0} > 0$ such that for $|z| > {R_0}$ the polynomial can be bounded below as
$$|P(z)| > \frac{1}{2}|{a_n}||z|^{n}.$$
How do I have to start this proof?
You can start with the triangle inequality:
$$|P(z)|\geq |a_n||z|^n-|a_{n-1}||z|^{n-1}-\cdots-|a_0|$$ so that $$|P(z)|\geq |a_n||z|^n\left(1-\dfrac{|a_{n-1}|}{|z|}-\cdots-\dfrac{|a_0|}{|z|^n}\right)$$ so we just need to show that $$\dfrac{|a_{n-1}|}{|z|}+\cdots+\dfrac{|a_0|}{|z|^n}\leq\dfrac{1}{2}$$ if $|z|$ is sufficiently large. But this holds since $$\lim_{|z|\rightarrow\infty}\dfrac{|a_{n-1}|}{|z|}+\cdots+\dfrac{|a_0|}{|z|^n}=0$$ You might want to formalize that last bit with $\varepsilon$ and $R_0$ depending how rigorous your proof needs to be.