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:
$$(a - b)\cdot(x-b)<0 \implies (a - b)\cdot(b-x)>0$$ $$\implies \text{by Cauchy-Schwarz} \mid a - b \mid \mid b-x \mid > 0$$ $$\implies \mid a - x + x - b \mid \mid b-x \mid > 0$$ $$\implies \text{using triangle inequality} \left(\mid a - x \mid + \mid x - b \mid \right) \mid b-x \mid > 0$$ $$\implies \mid a - x \mid > - \mid b-x \mid$$
I shouldn't have the negative sign, but can't see where I am going wrong. Some advice would be appreciated. Thank you.
Note that \begin{align}(a-x)\cdot(b-x)-(b-x)\cdot (b-x)&=\big((a-x)-(b-x)\big)\cdot(b-x)\\&=(a-b)\cdot(b-x)\\&=-(a-b)\cdot(x-b)>0.\end{align} Therefore, $$(a-x)\cdot (b-x)>(b-x)\cdot(b-x)=\|b-x\|^2.$$ By C-S, $$\|a-x\|\|b-x\|\ge (a-x)\cdot (b-x).$$ Therefore, $$\|a-x\|\|b-x\|>\|b-x\|^2.$$ Since $\|b-x\|>0$, we can divide both sides of the inequality above by $\|b-x\|$ to get $$\|a-x\|>\|b-x\|.$$