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$x+y\ge x+y$

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.

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    $\begingroup$ 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
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$x+y\ge x+y\quad \ldots\quad$ 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 $\,1 \gt 2\,$.

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