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$ ?

$|x-1|<1\implies |x^2-4x+3|=|(x-1)^2+2(1-x)|\leq $ $\leq |(x-1)^2|+|2(1-x)|<1^2+2\cdot 1=3.$
