Question

Prove that for all integers n > or = 1 and all real numbers x subscript 1, x subscript 2,...., x subscript n. absolute value of (x subscript 1+ x subscript 2+...+ x subscript n) <or= abs(x subscript 1) + abs( x subscript 2)+..+ abs(x subscript n)

I know how to do the triangle inequality where I just have 2 values in the reals. But I don't understand how to generalize it to more than 2 values in the reals.

Answer 1

we want to show that

|x1 + x2 + ... + xn| <= |x1| + |x2| + ... + |xn|

Let's take n=3

We know that the inequality holds for n=2, so now we have

|(x1+x2)+x3| <= |(x1+x2)| + |x3|
But, we already know that |x1+x2| <= |x1|+|x2|, so we now have

|x1+x2+x3| <= |x1|+|x2|+|x3|
as was desired.

You can see that using the associative property of real numbers extends the proof to any positive integer n.