Inequality determine minimum I want to show that a continuous function $f:[0, \infty) \to \mathbb{R}$ satisfying $f(x) \ge 2+x$ must has a minimum.
I want to use the extreme value theorem but i don't know how to rewrite the interval into a closed and bounded one in order to satisfy all the requirements of the Extreme Value Theorem. 
 A: You said:

I want to use the extreme value theorem but i don't know how to rewrite the interval into a closed and bounded one in order to satisfy all the requirements of the Extreme Value Theorem.

The idea is to use the extreme value theorem on a "suitable"
closed and bounded interval, and then show that the minimum of 
$f$ on this interval is in fact the overall minimum of $f$.
$f(0) \ge 2 + 0 > 0$, therefore we can define the interval $I$ as 
$$
 I = [0, f(0)] \, .
$$
$I$ is a closed, bounded interval and $f$ is continuous, therefore
$f$ has a minimum on $I$, i.e. there is a $x_0 \in I$ such that
$$ \tag{*}
  f(x_0) \le f(x) \text{ for all } x \in I \, . 
$$
In particular, $f(x_0) \le f(0)$.
Now for arbitrary $x \ge 0$, we have either
$$
 0 \le x \le f(0) \Longrightarrow f(x) \ge f(x_0)
$$
because of $(*)$, or
$$
 x > f(0) \Longrightarrow f(x) \ge 2 + x > 2 + f(0) > f(0) \ge f(x_0) \, .
$$
Therefore $f(x_0)$ is the overall minimum of $f$.
A: Let $m=\inf_{x\in[0,\infty)}f(x)$. Of course, $m\geq 2$. If $x>m$, then $f(x)> m+2$, so any minimizing sequence must eventually lie in $[0,m]$.
But $[0,m]$ is compact, and hence the infimum is attained (ie, there is a minimum) at some point inside it.
