Prove $ \left|x-1\right|<1$ implies $ \left|x^2-4x+3\right|<3$.

Question

Question:

Let $ x$ be a real number. Prove that if $ \left|x-1\right|<1$ then $ \left|x^2-4x+3\right|<3$.

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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$ ?

Answer 1

Now you have $$-1 < x-1 < 1, \text{ and thus also } -3<x-3 < -1$$

So it is ensured that

$$|x^2-4x+3|=|(x-3)(x-1)| < 3$$

Answer 2

By the triangle inequality we obtain: $$|x^2-4x+3|=|x-1|\cdot|x-3|\leq|x-3|\leq|x-1|+|-2|<3.$$

The triangle inequality is the following: $$|x+y|\leq|x|+|y|$$

Answer 3

Per your question "how can I show that |x−3|<3 ? ", you just use the triangle inequality, and the given that $|x-1| < 1$, you got: $|x-3| = |x-1+(-2)| \le |x-1|+|-2| < 1 + 2 = 3$

Answer 4

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 $.

Answer 5

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