What am I doing wrong here?

I am trying to show $\mid a-x|>|b-x|$ using the fact that $(a-b)\cdot (x-b)<0$ for $a,b,x\in {\mathbb{R}}^{N}$

So far, I have done this:

$(a-b)\cdot (x-b)<0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(a-b)\cdot (b-x)>0$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{by Cauchy-Schwarz}\mid a-b\mid \mid b-x\mid >0$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mid a-x+x-b\mid \mid b-x\mid >0$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{using triangle inequality}(\mid a-x\mid +\mid x-b\mid )\mid b-x\mid >0$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mid a-x\mid >-\mid b-x\mid $

I shouldn't have the negative sign, but can't see where I am going wrong. Some advice would be appreciated. Thank you.

I am trying to show $\mid a-x|>|b-x|$ using the fact that $(a-b)\cdot (x-b)<0$ for $a,b,x\in {\mathbb{R}}^{N}$

So far, I have done this:

$(a-b)\cdot (x-b)<0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(a-b)\cdot (b-x)>0$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{by Cauchy-Schwarz}\mid a-b\mid \mid b-x\mid >0$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mid a-x+x-b\mid \mid b-x\mid >0$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{using triangle inequality}(\mid a-x\mid +\mid x-b\mid )\mid b-x\mid >0$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mid a-x\mid >-\mid b-x\mid $

I shouldn't have the negative sign, but can't see where I am going wrong. Some advice would be appreciated. Thank you.