Let $p:\mathbb{R}\rightarrow{\mathbb{R}}$ be an everywhere positive polynomial, such that $p(x)>0$ for every $x\in\mathbb{R}$. Prove that there exists a $\eta>0$ such that $p(x)\ge{\eta}$ for every $x\in\mathbb{R}$.
ive attached a screenshot of the proof I'm trying to follow, but don't understand the section I've underlined in red. I don't see how this follows. The rest I understand however.
If $\lvert x\rvert\geqslant M$, then\begin{align}p(x)&=x^n\left(a_n+\frac{a_{n-1}}x+\cdots+\frac{a_1}{x^{n-1}}+\frac{a_0}{x^n}\right)\\&\geqslant x^n\left(a_n-\frac12\lvert a_n\rvert\right)\\&=x^n\left(a_n-\frac12a_n\right)\\&=\frac12a_nx^n\\&\geqslant\frac12a_nM^n.\end{align}