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 a x | > | b x | using the fact that ( a b ) ( x b ) < 0 for a , b , x R N
So far, I have done this:
( a b ) ( x b ) < 0 ( a b ) ( b x ) > 0
by Cauchy-Schwarz a b ∣∣ b x ∣> 0
a x + x b ∣∣ b x ∣> 0
using triangle inequality ( a x + x b ) b x ∣> 0
a x ∣> b x
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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Answers (1)

LitikoIDu6
Answered 2022-07-17 Author has 10 answers
Note that
( a x ) ( b x ) ( b x ) ( b x ) = ( ( a x ) ( b x ) ) ( b x ) = ( a b ) ( b x ) = ( a b ) ( x b ) > 0.
Therefore,
( a x ) ( b x ) > ( b x ) ( b x ) = b x 2 .
By C-S,
a x b x ( a x ) ( 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 .
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