# Triangle inequality proof in Spivak's calculus

1. Jan 6, 2014

### chemistry1

So hi, there's one little thing which I'm not understanding in the proof. After the inequality Spivak considers the two expressions to be equal. Why?!?

I just don't see why we can't continue with the inequality and when we have factorized the identity to (|a|+|b|)^2 we can just replace (a+b)^2 with (|a+b|)^2 and take the square root of both sides to finally have :

|a+b| <= |a|+|b|

Thank you for explaining !

2. Jan 6, 2014

### chemistry1

No need to answer, it's understood !

3. Jan 6, 2014

### FeDeX_LaTeX

You could also start from $-|a| \leq a \leq |a|$.

4. Jan 6, 2014

### chemistry1

Well, I have another question. When Spivak justifies the passage from the squares to |a+b| <= |a|+|b| he says the following : x^2<y^2 supposes that x<y for x,y in N. Now, the only thing bugging me is the following : Why didn't he do the following x^2<=y^2 supposes that x<=y for x,y in N ? Because what he says only justifies the inequality and not the equality ! Like a part is missing ! Am I right ? Thank you!