 # Proving ∣a−x|>|b−x| implies (a−b) * (x−b)<0 where a,b,x in RR^N Talon Mcbride 2022-07-16 Answered
What am I doing wrong here?
I am trying to show $\mid a-x|>|b-x|$ using the fact that $\left(a-b\right)\cdot \left(x-b\right)<0$ for $a,b,x\in {\mathbb{R}}^{N}$
So far, I have done this:
$\left(a-b\right)\cdot \left(x-b\right)<0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(a-b\right)\cdot \left(b-x\right)>0$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\text{by Cauchy-Schwarz}\mid a-b\mid \mid b-x\mid >0$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mid a-x+x-b\mid \mid b-x\mid >0$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\text{using triangle inequality}\left(\mid a-x\mid +\mid x-b\mid \right)\mid b-x\mid >0$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mid a-x\mid >-\mid b-x\mid$
I shouldn't have the negative sign, but can't see where I am going wrong. Some advice would be appreciated. Thank you.
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Note that
$\begin{array}{rl}\left(a-x\right)\cdot \left(b-x\right)-\left(b-x\right)\cdot \left(b-x\right)& =\left(\left(a-x\right)-\left(b-x\right)\right)\cdot \left(b-x\right)\\ & =\left(a-b\right)\cdot \left(b-x\right)\\ & =-\left(a-b\right)\cdot \left(x-b\right)>0.\end{array}$
Therefore,
$\left(a-x\right)\cdot \left(b-x\right)>\left(b-x\right)\cdot \left(b-x\right)=‖b-x{‖}^{2}.$
By C-S,
$‖a-x‖‖b-x‖\ge \left(a-x\right)\cdot \left(b-x\right).$
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
$‖a-x‖>‖b-x‖.$