# Triangle Inequality Proof

1. Oct 27, 2007

### Howers

To make it clear, I understand the theorem and several proofs of this theorem but the most basic one is not making sense.

Thm: |a+b|<or=|a|+|b|

Proof:
(a+b)^2 = a^2+2ab+b^2 < or = |a|^2 + 2|a||b|+|b|^2 = (|a| + |b|)^2
Taking the square root of both sides and remember that |x|=square root of (x^2), we can prove that |a+b| < or = |a| + |b| (Triangle inequality)

MY QUESTION: Why can you say that a^2+2ab+b^2 < or = |a|^2 + 2|a||b|+|b|^2 is true? Intuitively, this makes sense because if a or b is negative then obviously their product will make the left side less. But using this logic, why not just say that the triangle inequality is likewise intuitevly obvious? Obviously if one is negative, their sum must be less. So my question is, how do you rigoursly conclude ab<or=|a||b|.
This is the Cauchy inequality, but because it requires calculus to prove it does not seem logical as calculus relies on this very inequality!!

2. Oct 28, 2007

### d_leet

Well trivially x2=|x|2 for all x. So to show that a^2+2ab+b^2 is less than or equal to |a|^2+2|a||b|+|b|^2, you really only need to show that ab is less than or equal to |a||b|, you will have three cases, both are positive, both are negative, or one of a and b is negative.

3. Oct 28, 2007

### Gib Z

Cauchy-Schwarz does not require calculus to prove.