Let $p: \mathbb{R} \rightarrow \mathbb{R}$ be polynomial $p(t) = a_0 + a_1 t+ \cdots + a_n t^n $ $(a_n \neq 0)$.
I'd like to prove the following statement by induction: $\forall K>0$ there exists $r_k$ such that $|t|\geq r_k \Rightarrow |p(t)|\geq K$.
My attempt
Base case: For $n=1$, we have $p(t) = a_1t + a_0 $, $a_1 \neq 0$ and: $|p(t)| = | a_1t + a_0| \geq |a_1||t| - |a_0| \geq K \iff |t| \geq \frac{K+|a_0|}{|a_1|}$
Take $r_k = \frac{K+|a_0|}{|a_1|} $ and we have $|t|\geq r_k \Rightarrow |p(t)|\geq K$. Since $K$ was arbitrary, the statement holds for $n=1$.
Inductive step: Assume that the statement holds for $n-1$.
It means that given $ q(t) = a_0 + a_1 t+ \cdots + a_{n-1} t^{n-1} $, $(a_{n-1} \neq 0)$ and $K>0$, there exists $r_k $ such that $|t| \geq r_k \Rightarrow|q(t)| \geq K$.
Given $p(t) = a_0 + a_1 t+ \cdots + a_n t^n $, $(a_n \neq 0)$ , we can write : $p(t) = a_0 + t(a_1 + \cdots + a_n t^{n-1}) = a_0 + tq(t)$ .
Note that $q(t) = a_1 + \cdots + a_n t^{n-1} $ satisfies the induction hypothesis.
Observe that:
$$|p(t)| \geq |t||q(t)| -|a_0| \geq K \iff |t||q(t)| \geq K + |a_0| $$
Since $q(t)$ satisfies the induction hypothesis, there exists $r_1$ such that $|t|\geq r_1 \Rightarrow |q(t)| \geq \sqrt{K+|a_0|} $.
$r(t) = t$ also satisfies the induction hypothesis then there exists $r_2$ such that $|t|\geq r_2 \Rightarrow |r(t)| \geq \sqrt{K+|a_0|}$.
Take $r_k = \max \{r_1,r_2\}$.
Then $|t|\geq r_k \Rightarrow |p(t)|\geq K$. Thus the statement holds for $n$.
Am I right?
I think your attempt is correct. I suppose another way that use the Theorem Fundamental of Algebra, which implies that every real polynomial can be written by product of $(x-a)$ and $(x^2-ax+b)$. These polynomials is obviously tend ver infinite when x tend ver infinite