How to prove $|x+x_1+...+x_n| \ge |x|-(|x_1|+...+|x_n|)$?

Question

I'd like to return to college, which I had to leave due to money constraints, but now I'm preparing myself for real analysis. In my exercise book, there is a problem for showing an inequality:

$\forall x,x_1,...,x_n \in \mathbb{R}; n \in \mathbb{N}: |x+x_1+...+x_n| \geq |x|-(|x_1|+...+|x_n|)$

I have spent yesterday's evening, trying to somehow come up with an elegant solution, using the following inequalities I know to be true: $-|x|\leq x \leq |x|$; $|x+y| \leq |x| + |y|$; $|x-y| \geq ||x|-|y||$.

My latest strategy has been to show that $|x|-|x_1|\leq |x-x_1|$ and subsequently $|x-x_1| \leq |x+x_1|$, the following $x_2, ..., x_n$ would then be easy to prove. However, the second inequality doesn't even hold and I'm stuck.

Answer 1

Hint. If $y = x_1+x_2+\dots+x_n$, we have $$|x+y| \geqslant |x| - |y|$$ $$|x_1|+|x_2|+\dots+|x_n| \geqslant |y|$$ both by triangle inequality.

Answer 2

You go by induction. First prove that for $x,x_1$ it holds $|x+x_1| \geq |x|- |x_1|$, and this is simply the triangle inequality (or reversed triangle inequality if you like). Then consider $x, x_1, \dots,x_n, x_{n+1}$ assuming you have the thesis for $n$ numbers instead of $n+1$.

Consider $x, x_1, \dots, x_{n}+x_{n+1}$. Now apply the inductive hypothesis to get

$$ |x+x_1 + x_2 + \dots x_{n+1}| \geq |x|- (|x_1| + \dots + |x_n + x_{n+1}|) \geq\\ \geq |x|- (|x_1| + \dots + |x_n| + |x_{n+1}|)$$

by the triangle inequality applied to $|x_n + x_{n+1}|$.

Good luck with your comeback!