Need help with this! $\lvert x-y \rvert\lt\epsilon^2 \Rightarrow \lvert\,\lvert\sqrt x \rvert - \lvert \sqrt y \rvert \,\rvert \lt \epsilon$ Given $\Bbb R$, $ \epsilon \gt 0$. Prove that any positive $\Bbb R$
$$\lvert x-y \rvert\lt\epsilon^2 \Rightarrow \bigl\lvert\lvert \sqrt x \rvert - \lvert \sqrt y \rvert \bigr\vert \lt \epsilon$$
I'm trying to get this done all and I get is an dead end, such as squaring everyone up
$\left( \ \vert \ x-y \ \vert \ \right)^2\lt\left( \ \epsilon^2 \ \right)^2 \Rightarrow \\$
$\left( \  x-y \ \right)^2\lt\ \epsilon^4 \Rightarrow \\$
$x^2 - 2xy + y^2\lt\ \epsilon^4$
and that's pretty much it, from here can't go anywhere close to the objective.
Then again tried another aproach which led me to another dead end.
$$\vert \ x-y \ \vert \lt\epsilon^2$$
$$\Rightarrow u = x-y$$
$$\vert \ u \ \vert \lt \epsilon^2$$
$$\Rightarrow -\left( \epsilon^2 \right) \lt u \lt \epsilon^2$$
$$\Rightarrow \epsilon \gt 0$$
$$u \lt \epsilon^2$$
$$x-y\lt\epsilon^2$$
$$\sqrt {x-y}\lt\epsilon$$
$$\left(x-y\right)^{1/2}\lt\epsilon$$
After this I coudn't go any further and I ran out of any ideas, would anyone be kind to show show a proper solution?
 A: Whenever I see something like $a^2 - b^2$ in these kind of problems, I often try to see if factoring it into $(a-b)(a+b)$ is fruitful.
In this problem, we do not have $x^2 - y^2$, but we do have $|x - y|$ and also $|\sqrt{x} - \sqrt{y}|$, which suggested to me that using $|x - y| = |(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})|$ might be fruitful.
In this problem, if $x$ and $y$ are both nonnegative, then:
\begin{align}
||\sqrt{x}| - |\sqrt{y}|| &= |\sqrt{x} - \sqrt{y}| \ge \epsilon \\
& \implies \max(\sqrt{x}, \sqrt{y}) \ge \epsilon \label{max}\\
& \implies \sqrt{x} + \sqrt{y} \ge \epsilon \\
& \implies |\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| \ge \epsilon^2 \\
&\implies |x - y| \ge \epsilon^2
\end{align}
Therefore $|x - y| < \epsilon^2 \implies |\sqrt{x} - \sqrt{y}| < \epsilon$
Edit to add more details, as requested:
Since $\sqrt{x}$ and $\sqrt{y}$ are both greater than or equal to zero, and the distance between them is at least $\epsilon$, then we know that $\max(\sqrt{x}, \sqrt{y}) \ge \epsilon $. Otherwise, they would both be in the interval $[0, \epsilon)$, which would make it impossible for the distance between them to be at least $\epsilon$.
Then, since they are both nonnegative, we know that $\sqrt{x} + \sqrt{y} \ge \max(\sqrt{x}, \sqrt{y}) $, hence $\sqrt{x} + \sqrt{y} \ge \epsilon$.
So we started with assumption that $|\sqrt{x} - \sqrt{y}| \ge \epsilon$, and deduced (using the nonnegativity of $x$ and $y$) that $\sqrt{x} + \sqrt{y} \ge \epsilon$, which combine to give:
$$|\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| \ge \epsilon^2$$
Since $|\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| = |x-y|$, that means that we have shown that:
$$|\sqrt{x} - \sqrt{y}| \ge \epsilon \implies |x - y| \ge \epsilon^2$$
Therefore, we have also shown the contrapositive:
$$|x - y| < \epsilon^2 \implies |\sqrt{x} - \sqrt{y}| < \epsilon $$
A: Let us suppose that $0\le x< y$. Let $\theta = \frac{x}{y}\in [0, 1)$. We have
\begin{equation}
\begin{array}{cl}
\cr
&\theta \le \sqrt{\theta}\cr
\Rightarrow &-2 \sqrt{\theta} \le -2\theta\cr
\Rightarrow &1 - 2\sqrt{\theta} + \theta \le 1 - 2 \theta + \theta\cr
\Rightarrow &(1 - \sqrt{\theta})^2\le 1 -\theta\cr
\Rightarrow &(1 - \sqrt{\frac{x}{y}})^2\le 1 -\frac{x}{y}\cr
\Rightarrow &\sqrt{y} - \sqrt{x}\le \sqrt{y-x}
\end{array}
\end{equation}
When $0\le y< x$, we just swap $x$ and $y$ to get $\sqrt{x} -\sqrt{y}\le \sqrt{x - y}$. Finally, if $|x-y|<\epsilon^2$, we have
\begin{equation}
|\sqrt{x} - \sqrt{y}|\le \sqrt{|x-y|}<\sqrt{\epsilon^2}=\epsilon
\end{equation}
