Let $T : E \longrightarrow F$ a linear bijective operator between normed spaces, then $T$ is an isomorphism if and only if there are $c,k>0$ such that $$c||x|| \leq ||T(x)|| \leq k||x||,$$ for all $x \in E$.
My attempt was:
$(\Leftarrow)$ If there is $k>0$ such that $||T(x)||\leq k||x||$, then $T$ is continuous. So, by the fact that $T\in \mathcal{L}(E,F)$ and exists $c>0$ such that $||T(x)||\geq c||x||$, for all $x\in E$, then the inverse operator $T^{-1}$ exists and it's continuous. So, $T$ is an isomorphism, since $T$ is already bijective.
$(\Rightarrow)$ Supose that $T$ is an isomorphism. So $T$ is continuous and bijective. Thus exists $k>0$ such that $||T(x)||\leq k||x||$, for all $x\in E$.
BUT I DON'T KNOW HOW TO PROCEED.
I think the first inequality is given from the continuity of $T^{-1}$, but I don't know how to do it. Does someone know how to do it?