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.
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.
To prove the inequality for the sum of absolute values of n terms, we can use mathematical induction.
Step 1: Base case
Let's start with n = 1. In this case, we have only one real number, x₁. It can be written as:
| x₁ | ≤ | x₁ |
This is obviously true.
Step 2: Inductive hypothesis
Assume that the inequality holds for some arbitrary positive integer k:
| x₁ + x₂ + ... + x_k | ≤ | x₁ | + | x₂ | + ... + | x_k |
Step 3: Inductive step
Now, we need to prove that the inequality also holds for k + 1, i.e., we want to show:
| x₁ + x₂ + ... + x_k + x_(k+1) | ≤ | x₁ | + | x₂ | + ... + | x_k | + | x_(k+1) |
We can rewrite the left side of the inequality as:
| (x₁ + x₂ + ... + x_k) + x_(k+1) |
Now, using the triangle inequality for two real numbers, we can write:
| (x₁ + x₂ + ... + x_k) + x_(k+1) | ≤ | x₁ + x₂ + ... + x_k | + | x_(k+1) |
By our assumption (inductive hypothesis), we know that:
| x₁ + x₂ + ... + x_k | ≤ | x₁ | + | x₂ | + ... + | x_k |
Plugging this into our inequality, we get:
| (x₁ + x₂ + ... + x_k) + x_(k+1) | ≤ | x₁ | + | x₂ | + ... + | x_k | + | x_(k+1) |
Thus, the inequality holds for k+1 as well.
By the principle of mathematical induction, we have proven that the inequality holds for all positive integers n, as required.
This proof shows that the sum of the absolute values of n terms is always less than or equal to the sum of the absolute values of those individual terms.