Using Supremum to prove $|x+y| \leq |x|+|y|$ vs. my proof

Question

I saw postings on MathStackExchange for this but I need to prove it using the definition of Supremum. I proved it the way I see it. I do not see the Supremum way. I would appreciate help.

$ \forall x, y \ \ \ \epsilon \ \Re$

Prove $\lvert x+y\rvert \leq \lvert x\rvert + \lvert y\rvert$

My Proof:

$$ 1) \,(x+y)^2 = x^2 + 2xy + y^2 $$

Using the fact* that: $${\sqrt {x^2}} = \lvert x\rvert$$ $$x^2 = \lvert x\rvert ^2$$

We have$$2) \,x^2 + 2xy + y^2 = \lvert x\rvert ^2 + 2xy + \lvert y\rvert ^2 $$

And using: $$xy \leq \lvert xy\rvert$$

We have $$3) \,x^2 + 2xy + y^2 = \lvert x\rvert ^2 + 2xy + \lvert y\rvert ^2 \leq \lvert x\rvert ^2 + 2\lvert xy\rvert + \lvert y\rvert ^2 $$

Again using the fact*

$$\lvert x+y \rvert^2 =(x+y)^2$$ We have $$4) \,\lvert x+y \rvert ^2\leq \lvert x\rvert ^2 + 2\lvert xy\rvert + \lvert y\rvert ^2 $$

$$5) \,\lvert x+y \rvert^2 \leq (\lvert x\rvert + \lvert y\rvert )^2$$

$$6) \,\lvert x+y \rvert\leq \lvert x\rvert + \lvert y\rvert$$

Thanks in Advance for your help!

Answer 1

Your proof is correct. But of course one could avoid considering quadratic expressions. In any case nobody in his right mind would use a supremum argument to prove the triangle inequality. Note that this inequality already holds in ${\mathbb Z}$, which is centuries away from the idea of supremum.