0
$\begingroup$

$x+y\ge x+y$

Can I write this? Even though the left side will never be bigger than the right side is this inequality true? Got something similar on another task and wanted to know if I could write this as an answer.

$|a-b|\ge ||a|-|b||$ my task is to prove this by using the triangle inequality on $a=a-b+b$ and I've got to this point $|a-b|\ge |a|-|b|$ and I was also wondering if I could take the absolute value on both sides of the inequality to get to the other inequality.

$\endgroup$
  • 1
    $\begingroup$ Perhaps you could use the same argument you used to get $|a-b|\ge|a|-|b|$ on the expression $|b-a|$? $\endgroup$ – John Wayland Bales Aug 20 '18 at 23:15
2
$\begingroup$

$x+y\ge x+y\quad \ldots\quad$ Can I write this?

Yes, if $\,a=b\,$ then both $\,a \ge b\,$ and $\,a \le b\,$ hold true.

I've got to this point $|a-b|\ge |a|-|b|$

Since you got this for arbitrary $\,a,b\,$ the same inequality holds if you swap $\,a\,$ and $\,b\,$: $\,|b-a| \ge |b|-|a|\,$.

But $\,|a-b|=|b-a|\,$, so you have $\begin{cases}|a-b| \ge |a|-|b| \\ |a-b|\ge |b|-|a|\end{cases}\implies |a-b| \ge \big||a|-|b|\big|\,$, where the implication follows because $\,|x|\,$ is either $\,x\,$ or $\,-x\,$, so $\,\big||a|-|b|\big|\,$ is either $\,|a|-|b|\,$ or $\,|b|-|a|\,$.

was also wondering if I could take the absolute value on both sides of the inequality

Not in general, consider for example that $\,1 \gt -2\,$, but taking absolute values gives the (false) inequality $\,1 \gt 2\,$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.