Q: How can I prove the lower bound theorem? The proof for the upper bound theorem, that I've seen, goes like this:

A more elegant proof for the upper bound (the same logic can follow
with the lower) is to notice that if the quotient $q(x)$ has only
positive coefficients, all numbers greater than $a$ will have a positive
output (and thus never cross the $x$-axis).
$$f(x)=(x−a)q(x)+r(x)$$
For some $c>a$
$$f(c)=(c−a)q(c)+r(c)$$
$$\text{where}$$
$$c>a>0,$$
$$q(c)>0,$$
$$c−a>0,$$
$$r(c)>0$$
Hence, $f(c)=(c−a)q(c)+r(c)>0$ which implies that any value $>a$ cannot be a root and is
thus an upper bound.

The problem is that they say that the same logic can follow for the lower bound theorem, but I can't find that anywhere. The book I'm using says the exact same thing as well but doesn't supply the proof for the lower bound version.
So my question is: What is the proof for the lower bound version?
