Assume that a cone K (in an ordered Banach space) is normal, that is if $0\leq x\leq y$ implies that $||x||\leq N||y||. $ The constant $N$ is called the normality constant and is independent of $x$ and $y$. I want to show that if $K$ is normal, then:
$$\frac{||x+y||}{||x||+||y||} \geq \mu >0\hspace{1mm} \forall x,y \in K, ||x||+||y||>0 $$
This seems simple enough, but I am struggling to get it correct. Here's my attempt:
$$ \frac{||x+y||}{||x||+||y||}=\frac{||x+y||}{||y||\big{(}\frac{||x||}{||y||}+1\big{)}} \geq\frac{||x+y||}{||y||(N+1)}$$
The last inequality follows since if $||x|| \leq N||y||$, then $\frac{||x||}{||y||}\leq N$ and thus $\frac{1}{\big{(}\frac{||x||}{||y||}+1\big{)}} \geq \frac{1}{N+1}$.
Further, we get too that:
$$ \frac{||x+y||}{||y||(N+1)} \geq \frac{||x+y||}{(N+1)(||x||+||y||)}$$
I want to claim that we're finished as we found a constant $\mu = \frac{1}{N+1}$ such that everything works. But I'm not so sure if this is correct.
Can someone please check this? If not correct, please feel free to suggest.
The definition for cone here is the usual definition.
In order to invoke $\|x\| \le N\,\|y\|$ you need $0 \le x \le y$, but you only have $x,y \ge 0$. Thus, you cannot use this inequality.
Moreover, you have shown $$ \frac{\|x +y \|}{\|x\|+\|y\|} \ge \frac1{N+1} \, \frac{\|x +y \|}{\|x\|+\|y\|}. $$ This inequality is equivalent to $\frac{\|x +y \|}{\|x\|+\|y\|} \ge 0$, i.e., it does not show your desired inequality.