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I am trying to prove the inequality, $|\sqrt{x} - \sqrt{y}| \leq \frac{|x - y|}{2\text{min}\{\sqrt{x}, \sqrt{y}\}}$, holds for $x,y > 0$, and then using that to show that the function $f : \mathbb{R}^{+} \to \mathbb{R}, x \mapsto \sqrt{x}$ is continuous. I'm not positive that my proof of the inequality is correct, and also whether my relation to the function is sufficient enough.

My proof:

Let $x \geq y$. Then, $|\sqrt{x} - \sqrt{y}| = \sqrt{x} - \sqrt{y}$, and $|x - y| = x - y$. Hence, \begin{align*} 2\text{min}\{\sqrt{x}, \sqrt{y}\} &= 2\Big(\frac{1}{2}\Big(\sqrt{x} + \sqrt{y} - |\sqrt{x} - \sqrt{y}|\Big)\Big) \\ &= \sqrt{x} + \sqrt{y} - \sqrt{x} + \sqrt{y} \\ &= 2\sqrt{y}. \end{align*} Thus, $\sqrt{x} - \sqrt{y} \leq \frac{x - y}{2\sqrt{y}}$. Consequently, \begin{equation*} 0 \leq \sqrt{x} - \sqrt{y} \Rightarrow \sqrt{y} \leq \sqrt{x} \Rightarrow y \leq x \Rightarrow \frac{y}{\sqrt{y}} \leq \frac{x}{\sqrt{y}} \Rightarrow \frac{y}{2\sqrt{y}} \leq \frac{x}{2\sqrt{y}} \Rightarrow 0 \leq \frac{x-y}{2\sqrt{y}}. \end{equation*}

Therefore, $0 \leq \sqrt{x} - \sqrt{y} \leq \frac{x - y}{2\sqrt{y}}$, for $x \geq y$.

Now let $y \geq y - x$. Then, $|\sqrt{x} - \sqrt{y}| = \sqrt{y} - \sqrt{x}$ and $|x - y| = y - x$. Hence, \begin{align*} 2\text{min}\{\sqrt{x}, \sqrt{y}\} &= 2\Big(\frac{1}{2}\Big(\sqrt{x} + \sqrt{y} - |\sqrt{x} - \sqrt{y}|\Big)\Big) \\ &= \sqrt{x} + \sqrt{y} - (\sqrt{y} - \sqrt{x}) \\ &= \sqrt{x} + \sqrt{y} - \sqrt{y} + \sqrt{x} \\ &= 2\sqrt{x}. \end{align*}

Thus, $\sqrt{y} - \sqrt{x} \leq \frac{y - x}{2\sqrt{x}}$. Consequently, \begin{equation*} 0 \leq \sqrt{y} - \sqrt{x} \Rightarrow \sqrt{x} \leq \sqrt{y} \Rightarrow x \leq y \Rightarrow \frac{x}{2\sqrt{x}} \leq \frac{y}{2\sqrt{x}} \Rightarrow 0 \leq \frac{x-y}{2\sqrt{x}}. \end{equation*} Therefore, $0 \leq \sqrt{x} - \sqrt{y} \leq \frac{|x - y|}{2\text{min}\{\sqrt{x}, \sqrt{y}\}}$, for $ y \geq x$.

Now, let there be a function $f : \mathbb{R}^{+} \to \mathbb{R}, x \mapsto \sqrt{x}$. This function is continuous for a $\sqrt{x} \in \mathbb{R}^{+}$ iff for each $\epsilon > 0$ there exists a $\delta > 0$ s.t. for all $\sqrt{y} \in \mathbb{R}^{+}$ one has $|\sqrt{x} - \sqrt{y}| < \delta$. Let $\delta = \frac{|x - y|}{2\text{min}\{\sqrt{x}, \sqrt{y}\}}$. Then $|\sqrt{x} - \sqrt{y}| < \frac{|x-y|}{2\text{min}\{\sqrt{x}, \sqrt{y}\}}$. We proved this is true for cases $x,y > 0, x \geq y$ and $y \geq x$. Then we can conclude that $|f(\sqrt{x}) - f(\sqrt{y})| < \epsilon$, where $\epsilon = \frac{|f(\sqrt{x}) - f(\sqrt{y})|}{2\text{min}\{f(\sqrt{x}), f(\sqrt{y})\}}.$

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    $\begingroup$ Simply, the original inequality is equivalent to $$\left|\sqrt{x}+\sqrt{y}\right|\geq 2\min(\sqrt{x},\sqrt{y})$$ that is trivial. $\endgroup$ – Jack D'Aurizio Oct 27 '16 at 7:43
  • $\begingroup$ That is a consequence of $$\frac{x-y}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}.$$ $\endgroup$ – Jack D'Aurizio Oct 27 '16 at 7:44
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    $\begingroup$ My proof: Let x≥y ... Hence, 2min{√x, √y} = ... You are overcomplicating it, starting here. Since it's been assumed that $x \ge y$ already, then $\min(\sqrt{x},\sqrt{y})=\sqrt{y}$ follows directly. And, since the problem is symmetric in $x,y$ you don't even need to consider the other case $y \ge x$ separately. $\endgroup$ – dxiv Oct 27 '16 at 7:48
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Or if $x$ and $y$ are not both $0$,

$$|\sqrt{x} - \sqrt{y}|= \frac{|\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| }{|\sqrt{x} + \sqrt{y}| }\\ = \frac{|x - y| }{|\sqrt{x} + \sqrt{y}| }.$$

Since, $x \mapsto \sqrt{x}$ is non-negative, we have $|\sqrt{x} + \sqrt{y}| = \sqrt{x} + \sqrt{y} \geqslant 2 \min(\sqrt{x},\sqrt{y}), $ and

$$|\sqrt{x} - \sqrt{y}| \leqslant \frac{|x - y|}{2 \min(\sqrt{x},\sqrt{y}) }.$$

A more expedient way to prove continuity of $f(x) = \sqrt{x}$ (at $x=0$ as well) is to use

$$|\sqrt{x} - \sqrt{y}|^2 \leqslant |\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| = |x - y|.$$

If $|x - y| < \delta = \epsilon^2$, then $|\sqrt{x} - \sqrt{y}| < \epsilon.$

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