<s>[INST] Prove $ \left|x-1\right|<1$ implies $ \left|x^2-4x+3\right|<3$. Question:
Let $ x$ be a real number. Prove that if $ \left|x-1\right|<1$ then $ \left|x^2-4x+3\right|<3$.

We can write $ \left|x^2-4x+3\right|<3 $ as $ |x-3||x-1| < 3$. If I start with $ \left|x-1\right|<1$, how can I show that $ |x-3| < 3$ ?
 [/INST] This is true for
complex numbers as well as
real numbers.
The reason is that,
if $a$, $b$, and $c$
are positive reals,
then
$b \lt c$
implies
$ab \lt ac$.
(Do you see why this is true?)
Since
$|x^2-4x+3|
= |x-1||x-3|
$,
if
$|x-1| < 1$,
then
$|x^2-4x+3|
\lt |x-3|
$.
So,
we are done
if we can show that
$|x-3| < 3$.
But,
applying the
triangle inequality
($|u+v| \le |u|+|v|$),
$|x-3|
=|x-1-2|
\le |x-1|+|-2|
\lt 1+2
=3
$.
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