Triangle inequality proof in Spivak's calculus

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Handwritten-style printed proof excerpt: (a+b)^2 = a^2 + 2ab + b^2, then ≤ a^2 + 2|ab| + b^2, with |ab| = |a||b| and |a|^2 = a^2, ending at (|a|+|b|)^2; text reads 'form 1; this proof is motivated by the observation that'.


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 !
 
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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!