Let $f$ be a real continuous function defined on $D=[0,+\infty)$, $f(x)\geq 0$ for all $x\in D$, and $\lim_{x\to+\infty}\:f(x)=+\infty$. Prove that $f(x)$ attains its minimum on $D$.
Idea for a solution
Consider the subinterval $[0,M]$, where $M>0$. By the Extreme Value Theorem, there exists $c\in[0,M]$ such that $f(c)\leq f(x)$ for all $x\in[0,M]$. Now, $f(c)=\inf f([0,M])$.
My thinking was that one could construct a sequence such that as $M$ grows large, $f(c)$ approaches the minimum of $f$. I am not sure how to do this, or if this is even the correct approach.
Best Answer
It is better to "reverse" your approach. Since $\lim_{x\to+\infty}\:f(x)=+\infty$, there exists $R>0$ such that $f(x)>f(0)$ for all $x>R$. Then, since $f$ is continuous in $[0,R]$, $$\min_{[0,R]} f(x)\leq f(0)\leq\inf_{(R,+\infty)} f(x).$$ Hence the minimum value of $f$ over $[0,+\infty)$ exists and it is equal to $\min_{[0,R]} f(x)$.
P.S. There is no need of the condition $f(x)\geq 0$ for all $x\in D$.