Show that $f$ is unbounded below 
Let $f:\Bbb R\to \Bbb R$ be continuous and satisfies $|f(x)|\ge |x| $ for all $x$.Also $f(x+y)=f(x)+f(y)$ for all $x,y$.Show that $f$ is bijective.

My try:
$f$ is injective ;since $f(x)=f(y)\implies f(x-y)=0\implies |x-y|\le 0\implies x=y$.
To show that $f$ is onto.
Every continuous injection is either strictly increasing or decreasing.Hence if I can show that $f$ is both unbounded above and below then by the IVP we can show that $f$ is surjective.
Now since $|f(x)|\ge |x|>x$ forall $x$ so $f$ is unbounded above.However I am failing to show that $f$ is unbounded below also.
Please help me out here.

Note that $f(0)=f(0)+f(0)$, so $f(0)=0$.
Also, $f(-x)+f(x)=f(0)=0$, so $f(-x)=-f(x)$.
Since you've already showed that $f$ is unbounded above, it follows that it is also unbounded below.
