<s>[INST] Show that $\sqrt{x^2 +1 }$ is uniformly continuous. $x \in \mathbb{R}$ Let $x,y \in \mathbb{R} $ such that;
If $|x| \leq \delta$ then $|y| = |x| +(|y|-|x|) \leq |x| +||y|-|x|| \leq |x| + |y-x| < 2 \delta$
$$|\sqrt{x^2 +1 } - \sqrt{y^2 +1 }| = \sqrt{y^2 +1 } - \sqrt{x^2 +1 } \leq \sqrt{y^2 +1 } - 1  \leq y^2 +1 -1 < 2\delta := \epsilon $$
Now, this is the part I am having trouble with
if $|x| > \delta$ 
$$|\sqrt{x^2 +1 } - \sqrt{y^2 +1 }| \leq |\frac{x^2 - y^2}{\sqrt{x^2 +1 } + \sqrt{y^2 +1 }}| \color{red}{\leq \frac{(x+y)^2}{2\sqrt{\delta^2 + 1}}}...$$
I don't think the red term makes much sense. As the $2xy$ term could very well be negative. 
Could someone suggest a next step please.
 [/INST] You came close. We have
$$\left|\sqrt{1+x^2}-\sqrt{1+y^2}\right|=\frac{|x^2-y^2|}{\sqrt{1+x^2}+\sqrt{1+y^2}} =|x-y|\frac{|x+y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}. $$
Now 
$$\frac{|x+y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}\le \frac{|x|+|y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}\lt 1. $$
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