In Epsilon-Delta proof, why is $3x^2|y|/(x^2 + y^2) \Leftarrow 3|y| = 3\sqrt{y^2} \Leftarrow 3\sqrt{x^2 + y^2} $ This is part of a Epsilon-Delta proof to show that: $$\lim_{(x, y) \to 0}\frac{3x^2y}{x^2 + y^2} = 0$$ The proof began with: $$\frac {3x^2|y|}{(x^2 + y^2)} < \epsilon$$ It was then pointed out that $x^2 \Leftarrow x^2 + y^2 \because y^2 >= 0$, which I understood. 
In that way, I get why: $$\frac {3x^2|y|}{(x^2 + y^2)} \le 3|y|$$ But I don't understand how they can say it is equal to: $$ 3\sqrt{y^2} \le 3\sqrt{x^2 + y^2} $$
 A: Assume $x,y\in\mathbb{R}$. First, you can start saying that $x^{2}\geq 0$, then adding $y^{2}$ in both sides of the inequality you get:
$$ x^{2}+y^{2}\geq y^{2} \Rightarrow \sqrt{x^{2}+y^{2}}\geq\sqrt{y^{2}}$$
because $f(t)=\sqrt{t}$ is monotone increasing on its domain. Now use that $g(t)=3t$ is also monotone increasing and then you have:
$$3\sqrt{x^{2}+y^{2}}\geq 3\sqrt{y^{2}}=3\vert{y}\vert,$$
this means
$$3\sqrt{x^{2}+y^{2}}\geq 3\vert{y}\vert,$$
as $(x,y)\to (0,0)$ you can suppose that $(x,y)\neq (0,0)$,  so  $$x^{2}+y^{2}\neq 0$$ and you can write:
$$3\vert{y}\vert\leq 3\sqrt{x^{2}+y^{2}}<3\delta.$$
Second, you can say that $y^{2}\geq 0$ then adding in both sides $x^{2}$ you get:
$$ y^{2}+x^{2}\geq x^{2} $$
we know that  $x^{2}+y^{2}>0$ then you can divide by $x^{2}+y^{2}$ and get:
$$ 1\geq \frac{x^{2}}{x^{2}+y^{2}},$$
Now you can multiply both sides off the last inequality by $\vert{y}\vert\geq 0$:
$$ \vert{y}\vert\geq \frac{x^{2}\vert{y}\vert}{x^{2}+y^{2}} $$
Multiply by 3 both sides to get:
$$ \frac{3x^{2}\vert{y}\vert}{x^{2}+y^{2}}\leq 3\vert{y}\vert<3\sqrt{x^{2}+y^{2}}<3\delta.$$
You can put $\delta(\epsilon)=\displaystyle{\frac{\epsilon}{3}}$. Notice that you always have:
$$y^{2}\geq 0 \Rightarrow x^{2}+y^{2} \geq x^{2} \Rightarrow \sqrt{x^{2}+y^{2}}\geq\vert{x}\vert$$
in the same way
$$x^{2}\geq 0 \Rightarrow x^{2}+y^{2} \geq y^{2} \Rightarrow \sqrt{x^{2}+y^{2}}\geq\vert{y}\vert$$
From the inequality $x^{2}+y^{2} \leq x^{2}$ you can directly multiply both sides by $3\vert{y}\vert$ to have: 
$$\frac{x^{2}}{x^{2}+y^{2}}\leq 1 \Rightarrow \frac{3\vert{y}\vert x^{2}}{x^{2}+y^{2}}\leq 3\vert{y}\vert$$
