You need to assume that $f$ is continuous; otherwise, there are counterexamples. You also need to specify the domain. Is it $(0,\infty)$? In this case, the statement isn't true.
If the domain is $[0,\infty)$:
Since $f(x)\ge x^2$, there is an $M>0$ such that $\tag{1}f(x )\ge f( 0)\ \text{ for all }\ x\ge M.$
Assuming $f$ is continuous, it does have a global minimum in the closed, bounded interval $[0,M]$.
By (1), this would also be the global minimum of $f$ in $[0,\infty)$.
Note that you just have to prove that there is a minimum of $f$, you don't have to explicitly find it (with the information given, this would be impossible to do).