Hello guys it would be a great help if you could look through this proof and point out any errors, I would like to know whether my proof is coherent, and whether it needs more/less detail.
We will be assuming that the general properties of $\mathbb{R}$ and the results of those properties hold, such as that the square root function is increasing, and that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ when $a,b $ are nonnegative.The definition of $|x|$ is as follows, for $x \geq 0, |x|=x,$ for $x<0, |x|=-x$.
Lemmas:
$|x| \geq x$ follows from definition.
We also note that $|x|=\sqrt{(x^2)}$. We prove this by cases. For $x \geq 0$ it is obvious that the equality holds. For $x<0$, since $-x>0$, $\sqrt{x^2}=\sqrt{(-x)(-x)}=-x=|x|$ . Hence the equality follows.
From the previous lemma it follows that $|x|^2=x^2$
It has to be true that $|xy|=|x||y|$. From the second lemma we know that $|xy|=\sqrt{(xy)^2}=\sqrt{x^2y^2}=\sqrt{x^2}\sqrt{y^2}=|x||y|$
The proof:
$(|a|+|b|)^2=|a|^2+2|a||b|+|b|^2=a^2+2|ab|+b^2 \geq a^2+ 2ab +b^2=(a+b)^2=(|a+b|)^2$
Given that the positive square root function is an increasing function, we take the square root of both sides(this can be done since both sides of the inequality are nonnegative) and obtain $|a|+|b| \geq |a+b|$. This concludes the proof.
