Let $(X, \|\cdot\|)$ be a Bench space. Assume that $x_n \in X$ for all $n=0,1,....$ We may notice that $\|f\|\geq \|g\|-\|f-g\| $ for $f, g\in X.$
Put $y= \sum_{n=0}^{\infty} x_n.$ Notice that
$\|y\| \geq \|x_1\| - \|y-x_1\| \geq \|x_1\|- \|x_0\|- \|\sum_{n=2}^{\infty} x_n\|$.
Now (using the fact that $s-t\geq s$ if $t\geq 0$), we may obtain that
$$\|y\|\geq \|x_1\|.$$
My question is: Where is the flaw in this proof?
I think there is a flaw in the proof. If so, under what conditions on $\|x_i\|$ we may conclude that $\|y\|\geq \|x_1\|$?
I think if we assume $\|x_1\|\geq \|x_0\|$ and $\|x_1\|\geq \|\sum_{n=2}^{\infty} x_n\|$ then we may conclude that $\|y\|\geq \|x_1\|$. But I do not know how. Thanks.
Since you are dealing with norms of elements, the inequalities you are dealing with are about numbers and have little to do with Banach spaces nor norms.
Say $\|x_1\|=5$, $\|x_0\|=5$, $\|\sum_2^\infty x_n\|=0$. Then you want to conclude, from $$ \|y\|\geq0=5-5-0=\|x_1\|-\|x_0\|-\left\|\sum_2^\infty x_n\right\|, $$ that $\|y\|\geq5$. For instance take $x_0=5$, $x_1=-5$, $x_2=x_3=\cdots=0$.