Compact operator and norm 
Let $E,F$ and $G$ be normed spaces, $f\in \mathcal{L}(E,F)$ and $g\in \mathcal{L}(F,G)$. Suppose that $g$ is injective and $f$ is compact. 
  Show that, $\forall \varepsilon > 0$, there exists $M>0$ such that:
  $$\|f(x)\| \leq \varepsilon \|x\| + M \|g(f(x))\|, \quad \forall x \in E.$$

$$\|f(x_n)\| >  \varepsilon \|x_n\| + M \|g(f(x_n))\|$$
My attempt:
Assume that the conclusion does not hold. Then there exists $\varepsilon > 0$
and $x_n$ such that
$$\|f(x_n)\| >  \varepsilon \|x_n\| + n \|g(f(x_n))\| \iff \|f(x_n)\| >  \varepsilon + n \|g(f(x_n))\|, \quad  \|x_n\|=1, \quad \forall n \in
\mathbb{N}.$$
Since $f$ is compact and $x_n$ is bounded, there exists a subsequence $(x_{n_k}) $
such that $f(x_{n_k}) \to y \in F$.
Thus:
$$\|f(x_{n_k})\| >  \varepsilon + n_k \|g(f(x_{n_k}))\|, \quad \forall k \in \mathbb{N}$$
If $g(f(x_{n_k}))$ is bounded, it's done. However, I couldn't prove that
$g(f(x_{n_k}))$ is bounded and I'm not sure if it holds.
Any hints? New ideas are welcome.
 A: Divide by $n_k$ in the inequality $$\tag{*}\|f(x_{n_k})\| >  \varepsilon + n_k \|g(f(x_{n_k}))\|, \quad k \in \mathbb{N}$$
and letting $k$ going to infinity, we get, by boundedness of $\left(f\left(x_{n_k}\right)\right)_{k\geqslant 1}$, that $\lim_{k\to +\infty}g\left(f\left(x_{n_k}\right)\right)=0$. By continuity of $g$ we also have that $\lim_{k\to +\infty}g\left(f\left(x_{n_k}\right)\right)=g(y)$ and using injectivity of $g$ we get $y=0$. Since $f$ is bounded, we have for $k$ large enough that $\left\lVert f\left(x_{n_k}\right)\right\rVert\leqslant \varepsilon $ hence by (*), we get a contradiction.
A: $\|f(x_{n_k})\|\leq \|x_{n_k}\| \|f\|=\|f\|$. This implie that from your inequality,
$\|f\|\geq \|f(x_{n_k})\|>\epsilon +n_k\|g(f(x_{n_k}))\|$ 
since $f(x_{n_k})$ converges towards $y$, $lim_n\|g(f(x_{n_k}))\|=\|g(y)\|$.
We deduce:
$\|f\|\geq \epsilon+n_k\|g(y)\|$. This implies that $g(y)=0$ (since $n_k\|g(f(x_{n_k}))\|$ has to be bounded) and $y=0$ since $g$ is injective. The equality again implies:
$lim_k \|f(x_{n_k})\|>\epsilon+ lim_{k}(n_k\|g(f(x_{n_k}))\|)$ which implies that $0>\epsilon$. Contradiction.
