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Show $h(x)=1/(x^2+1)$ satisfies the $\epsilon-\delta$ definition for uniform continuity on $\mathbb{R}$

Attampt:

Let $\epsilon>0$, there exists a $\delta=\epsilon/2$ such that $|x-y|<\delta$ where $x,y\in\mathbb{R}$.

$\begin{align*}|h(x)-h(y)|&=\left|\frac{1}{1+x^2}-\frac{1}{1+y^2}\right|\\&=\left|\frac{x^2-y^2}{(1+x^2)(1+y^2)}\right|\\&\leq|x-y|\left|\frac{x+y}{(1+x^2)(1+y^2)}\right| \\&<|x-y|\cdot\left|\frac{1+x^2+1+y^2}{(1+x^2)(1+y^2)}\right|\\&<|x-y||1+1|\\&<2\delta<\epsilon\end{align*}$

I found the same the question Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$. My problem is: I asked my instructor that I need to break the question into three cases $I_1=[-1,1],I_2=(-\infty,-1],I_3=[1,\infty)$ because $x+y<1+x^2+1+y^2$ is different in these three intervals,but for me $x+y<1+x^2+1+y^2$ is true for the entire real line. Can someone tell me why? Thanks in advanced.

Community
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Simple
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  • Well, you really need that $|x+y|\leq 1+x^2+1+y^2$. It's actually easier to show that $2(x+y)\leq 1+x^2+1+y^2$ and $-2(x+y)\leq 1+x^2+1+y^2$. – Thomas Andrews Nov 05 '15 at 16:18
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    I would just prove that $-1-u^2<2u<1+u^2$ and this gets you the result by adding this inequality for $u=x$ and $u=y$ – Thomas Andrews Nov 05 '15 at 16:20
  • @ThomasAndrews on $\mathbb{R}$ or $I_1,I_2,I_3$? – Simple Nov 05 '15 at 16:22
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    But you say, "But for me ... " And yet you don't offer a proof. If you can prove it for every pair $x,y$ without breaking it into cases, do so. (But the way, $(x,y)$ is a pair, so it is strange to talk about the domain being intervals.) – Thomas Andrews Nov 05 '15 at 16:23

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It's not clear what cases your instructor wants you to apply, since a pair $(x,y)$ is not inside an interval.

Here's how I'd prove it. First prove:

Lemma: If $u\geq 0$, then $1+u^2\geq u$.

Then apply this to $u=|x|$ and $u=|y|$, adding, we get:

$$1+x^2 + 1+y^2\geq |x|+|y| \geq |x+y|$$

The final inequality being the triangle inequality.

Thomas Andrews
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