understanding part of proof in Banach-Steinhaus theorem

Theorem: If a sequence of linear bounded operators $$\{A_n\}_{n=1}^{\infty}$$ is a Cauchy sequence in every point of the Banach space $$E_x$$, then the sequence of norms $$\{\lVert A_n \rVert\}_{n=1}^{\infty}$$ is bounded.

First, let the sequence $$\{A_n\}_{n=1}^{\infty}$$ be a Cauchy sequence in every point of the space $$E_x$$.

Lets assume the opposite, that the sequence of norms $$\{\lVert A_n \rVert\}_{n=1}^{\infty}$$ is not bounded. Then $$\{\lVert A_n \rVert\}_{n=1}^{\infty}$$ is not a bounded set for closed ball $$\lVert x-x_0 \rVert \le \epsilon$$.

Lets assume that $$\lVert A_n x \rVert \le C$$ for some constant $$C$$, for every n and for every x in the closed ball $$\overline{B}(x_0, \epsilon)$$.

Now, for every $$\xi\in E_x$$ the element

$$x=\frac{\epsilon}{\lVert \xi \rVert}\xi+x_0$$ belongs to that ball and therefore $$\lVert A_n x \rVert \le C$$ for $$\forall n \in \mathbf N$$.

From there we conclude that

$$\frac{\epsilon}{\lVert \xi \rVert}\lVert A_n\xi \rVert - \lVert A_n x_0 \rVert \le \lVert \frac{\epsilon}{\lVert \xi \rVert} A_n\xi + A_n x_0 \rVert \le C$$

I really don't understand the last inequality. I'm familiar with the triangle inequality and reversed triangle inequality, but I can't seem to convince myself of this one.

$$\frac {\epsilon} {\|\xi\|} \|A_n\xi\|= \|(\frac {\epsilon} {\|\xi\|}A_n\xi+A_nx_0) -A_nx_0\|\leq \|\frac {\epsilon} {\|\xi\|}A_n\xi+A_nx_0 \|+\|A_nx_0\| =\|A_nx\|+\|A_nx_0\|$$ which gives the inequality.