2
$\begingroup$

Prove the following inequality:

$$|x+x_1+x_2+...+x_n|\ge|x|-(|x_1|+|x_2|+...+|x_n|)$$

I've spent a long time on this one. I've come up with something but I don't think it's correct.

According to the triangle inequality $|x+y|\le|x|+|y|$:

$$|x_1+x_2+...+x_n|\le|x_1|+|x_2|+...+|x_n|$$

So then

$$|x|+|x_1|+|x_2|+...+|x_n|\ge|x|-(|x_1|+|x_2|+...+|x_n|)$$ $$\Downarrow$$ $$|x_1|+|x_2|+...+|x_n|\ge-|x_1|-|x_2|-...-|x_n|$$

Which is supposed to be proof of the first statement according to some people here but I think we only proved that a bigger number than the one in the L.H.S. is bigger than the one in the R.H.S., which doesn't mean that the L.H.S. is bigger than the R.H.S.

Thanks for your help.

$\endgroup$
2
$\begingroup$

Using the triangle inequality, $$|x|=|x+x_1+\cdots +x_n-(x_1+\cdots + x_n)|\le $$ $$\le|x+x_1+\cdots+x_n|+|-x_1|+\cdots +|-x_n|=$$ $$=|x+x_1+\cdots+x_n|+|x_1|+\cdots +|x_n|$$

$\endgroup$
  • $\begingroup$ Thanks a lot, that worked. But I'm curious, is the proof I proposed valid? (in my opinion it's not) $\endgroup$ – Martijon Nov 7 '16 at 21:27
  • $\begingroup$ It could be said that the proof that a>b that I proposed only states that another number (c) that is greater than b, c>b, is also greater than a, c>a, which doesn't imply that a>b. $\endgroup$ – Martijon Nov 7 '16 at 22:05

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.