- #1

- 445

- 3

__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!!