I'm starting school soon and doing some review problems to prep for Calculus. I'm a bit stuck on this problem:

Show that if $|x| < 4$ then $|x^2-2x+3| < 27$.

I know that I have to use the Triangle Inequality. Here is what I've written so far:

$$|x| <4 \implies |x|^2 <16$$


$$2|x| <8\ \text{ and } \ |3|=3$$

The triangle inequality gives: $|x^2-2x+3| \leq |x^2|+|-2x|+|3|$ So combining all the inequalities I came up with using the $|x| <4$ I get:

$$|x^2-2x+3|\leq 16+8+3$$

$$|x^2-2x+3|\leq 27$$

My problem is the $\leq$ sign. Really I'm looking to show that the < "less than" is true, not $\leq$ "less than or equal to". Can anyone shed some light on where I went wrong?

  • $\begingroup$ Equality is achieved when the sides lie on a line. Show that it cannot be achieved. $\endgroup$
    – MathGod
    Dec 13 '14 at 17:59

The signs $<$ and $\le$ combine into the $<$ sign, $$|x^2-2x+3|\le|x^2|+|{-}2x|+|3|<16+|{-}2x|+|3|<16+8+|3|=27$$ implies $$|x^2-2x+3|<27.$$

This is no different from things like $$a\le b<c\le d\implies a<d$$ which should be obvious.

  • $\begingroup$ Ahhh okay I see, thanks! If I'm understanding correctly - When I was doing the |x|<4 I was always using the less than sign, it could not have been equal to 4. Same goes with the other inequalities I reached that way, and so in the end when I combine them all there can be no possibility of equality either... $\endgroup$
    – nnnm
    Dec 13 '14 at 18:06
  • $\begingroup$ Yes.​​​​​​​​​​​ $\endgroup$ Dec 13 '14 at 18:44

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