$f(n) = O(g(n))$ implies $g(n) = O(f(n))$

Question

How do I prove/disprove $f(n) = O(g(n))$ implies $g(n) = O(f(n))$?

I got to $f(n) \le c * g(n)$ easily enough from the definition of Big O, but I'm not sure how to get to $c*f(n) \ge g(n)$.

Answer 1

NO. $f(x)=x, g(x)=x^2$ is a counterexample.