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!
Best Answer
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?