Show that $f$ attains its minimum in $\mathbb R$. Let $f: \mathbb R \to \mathbb R$ be a continuous function, such that $f(x)\ge |x|$ for all $x \in \mathbb R$. Show that the function $f$ attains its minimum in $\mathbb R$. 


My thoughts:
When $|x|$ is big, then $f(x)$ is also big, since it's continuous it reaches its minimum in $\mathbb R$ by extreme value theorem. How do I show this, and do I need to go in the specifics a bit more?
Let $f(0)=0$.

Then $f$ attains its minimum at $0$ because $f(x)\ge 0\forall x$

So ,we assume that Let $f(0)&gt;0 \forall x$.

Then consider the function $g(x)=\dfrac{1}{f(x)}$ which is also continuous.

Also $g(x)\le \dfrac{1}{|x|}$ forall $x\in \Bbb R$ and $g(0)&gt;0$.

So consider the interval $A=[-\dfrac{1}{g(0)},\dfrac{1}{g(0)}]$ which is compact ,So $g$ being bounded attains its maximum in $A$ say at the point $x_0$.
Hence $g(x)\le g(x_0)\forall x\in A$.

For $x\in A^c;|x|&gt;\dfrac{1}{g(0)}\implies g(x)\le \dfrac{1}{|x|}&lt;g(0)$.

Also $g(0)\le g(x_0) $ since $x_0\in A$. Hence $g(x)\le g(x_0)\forall x$

So $g$ attains its maximum and hence $f$ attains its minimum at $x_0$.
Let $f(0)=0$.

Then $f$ attains its minimum at $0$ because $f(x)\ge 0\forall x$

So ,we assume that Let $f(0)&gt;0 \forall x$.

Then consider the function $g(x)=\dfrac{1}{f(x)}$ which is also continuous.

Also $g(x)\le \dfrac{1}{|x|}$ forall $x\in \Bbb R$ and $g(0)&gt;0$.

So consider the interval $A=[-\dfrac{1}{g(0)},\dfrac{1}{g(0)}]$ which is compact ,So $g$ being bounded attains its maximum in $A$ say at the point $x_0$.
Hence $g(x)\le g(x_0)\forall x\in A$.

For $x\in A^c;|x|&gt;\dfrac{1}{g(0)}\implies g(x)\le \dfrac{1}{|x|}&lt;g(0)$.

Also $g(0)\le g(x_0) $ since $x_0\in A$. Hence $g(x)\le g(x_0)\forall x$

So $g$ attains its maximum and hence $f$ attains its minimum at $x_0$.