Proving an inequality containing 3 variables

Question

I am not really sure how to prove this inequality. I thought about using AM-GM for 3 variables ($\frac{a+b+c}{3}\ge abc$), but that didn't work out, so I then decided to try and use cases ($x\ge0$) and ($x<0$). However, I couldn't really get that far by doing so.

Here is the question

Let $a,b,c$ be three real numbers.
Prove that $|a-c| \leq |a-b|+|b-c|$

I have tried to simplify this expression by squaring both sides, but that just left me with a mess, and I am not sure what to do next.

Any help will be greatly appreciated!

Answer 1

The square of the left-hand side is $$(a-c)^2 = (a-b+b-c)^2 = (a-b)^2 + (b-c)^2 + 2 (a-b)(b-c).$$ The square of the right-hand side is $$(|a-b|+|b-c|)^2 = (a-b)^2 + (b-c)^2 + 2|a-b||b-c|.$$

So, all you need to do is show $(a-b)(b-c) \le |a-b||b-c|$. Can you finish the job?

Answer 2

Squaring of the both sides helps!

We need to prove that: $$a^2-2ab+b^2+b^2-2bc+c^2+2|(a-b)(b-c)|\geq a^2-2ac+c^2$$ or $$|(a-b)(b-c)|\geq(b-a)(b-c),$$ which is obvious.