Is there an extension of the reverse triangle inequality to $n$ variables? The triangle inequality for real numbers states $|x+y| \leq |x| + |y|$. This is extended easily on induction to the corresponding result for $n$ variables
$$|x_1 + x_2 + ... + x_n| \leq |x_1| + |x_2| + ... + |x_n|, \quad \text{i.e. } \quad \left| \sum_{i=1}^n x_i \right| \leq \sum_{i=1}^n |x_i|.$$
The two variable triangle inequality (and indeed the $n$ variable version) is 'as tight as possible' since equality can be achieved; this happens when either of the variables is zero or when both have the same sign.
The reverse triangle inequality goes in the opposite direction; it states $|x-y| \geq ||x| - |y||$. This inequality is also as tight as possible; again we have equality when either of the variables is zero or they both have the same sign.
To make the reverse triangle inequality look more like the normal triangle inequality, we can put $y \mapsto -y$ to give $|x+y| \geq ||x| - |y||$. This inequality remains as tight as possible, except now we have equality when either of the variables is zero or they have opposite signs. Hence in the two variable case we can say
$$||x| - |y|| \leq |x+y| \leq |x| + |y|.$$
It isn't obvious to me whether the reverse triangle inequality can be extended to 3 variables; trying to extend it in the same way one extends the normal traingle inequality doesn't seem to work, and I couldn't find anything about it online. My intuition tells we that there probably isn't an extension because subtract is not commutative.
If is it the case there is no such generalisation of the reverse triangle inequality, does there exist a different formula for a lower bound on $|x+y+z|$, or is the best we can do $0 \leq |x+y+z|$?
 A: First, you need to rearrange your numbers so that
$$
|x_1| \ge |x_2| \ge \cdots \ge |x_n|
$$
We will use the following inequality to prove your point,
$$
|x|+|y| \ge |x+y| \ge ||x|-|y||
$$
For $n = 3$, we know that
$$
|x_1+x_2+x_3| \ge ||x_1|-|x_2+x_3||
$$
If $|x_1| \ge |x_2+x_3|$, i.e., $ |x_1| \ge |x_2+x_3| \ge |x_2|+|x_3|$, then
$$
|x_1+x_2+x_3| \ge |x_1|-|x_2+x_3| \ge |x_1|-|x_2| - |x_3|,
$$
If $|x_1| \le |x_2+x_3|$, then
$$
|x_1+x_2+x_3| \ge |x_2+x_3|-|x_1| \ge ||x_2| - |x_3||-|x_1|,
$$
Since $|x_2| \ge |x_3|$, we can rewrite the above inequality as
$$
|x_1+x_2+x_3| \ge |x_2+x_3|-|x_1| \ge |x_2| - |x_3|-|x_1|,
$$
Combining the above results, we have
$$
|x_1+x_2+x_3| \ge \max( |x_1|-|x_2|,  |x_2| -|x_1| ) - |x_3|
$$
Since $|x_1| \ge |x_2|$, we can rewrite the above inequality as
$$
|x_1+x_2+x_3| \ge |x_1|-|x_2| - |x_3|
$$
For $n \ge 4$, we would obtain
$$
|x_1+\cdots+x_n| \ge |x_1|-\sum_{i=2}^{n}{ |x_i| }
$$
However, keeping the right-hand side term to be positive would be a very intricate task.
