Triangular inequality - sum of the absolutes and absolut of the sum

Question

I am wanting to verify (and proof) that

$$\sum|a-b| \ge |\sum(a) - \sum(b)| $$

I know, from triangular inequality, that

$$ \sum |a – b| \ge \sum(|a| – |b|) $$

And so

$$ \sum |a – b| \ge \sum |a| – \sum |b| $$

I am not being able to proof that

$$ \sum |a| - \sum |b| \ge | \sum (a - b) | $$

Answer 1

In fact, this $$\sum |a| - \sum |b| \ge | \sum (a - b) |$$ is not true as the quantity on the left side could be negative.

To prove

$$\left\vert\sum(a) - \sum(b)\right\vert\le \sum|a-b|$$ just note that $$\left\vert\sum(a) - \sum(b)\right\vert = \left\vert\sum(a -b)\right\vert = \big/t := a-b\big/ = \left\vert\sum t\right\vert \le \sum\left\vert t\right\vert = \sum|a-b|$$