Proof of series inequality using triangle inequality
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 $|x+y|\le|x|+|y|$:
$$|x_1+x_2+...+x_n|\le|x_1|+|x_2|+...+|x_n|$$
$$|x|+|x_1|+|x_2|+...+|x_n|\ge|x|-(|x_1|+|x_2|+...+|x_n|)$$ $$\Downarrow$$ $$|x_1|+|x_2|+...+|x_n|\ge-|x_1|-|x_2|-...-|x_n|$$
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.
calculus sequences-and-series inequality triangles
MartijonMartijon
Using the triangle inequality, $$|x|=|x+x_1+\cdots +x_n-(x_1+\cdots + x_n)|\le $$ $$\le|x+x_1+\cdots+x_n|+|-x_1|+\cdots +|-x_n|=$$ $$=|x+x_1+\cdots+x_n|+|x_1|+\cdots +|x_n|$$
vadim123vadim123
$\begingroup$ 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
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