<s>[INST] Meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$. I know that, for $|x|\leq 1$, $e^x$ can be bounded as follows:
\begin{equation*}
1+x \leq e^x \leq 1+x+x^2
\end{equation*}
Likewise, I want some meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$.
The first thing that comes to my mind is $\sqrt{a^2}-\sqrt{b} < \sqrt{a^2+b}$, but plugging this in ends up with a non-sense lower-bound of $-\sqrt{b}$ even though the target number is positive.
\begin{equation*}
\big(\sqrt{a^2}-\sqrt{b} \big) - a < \sqrt{a^2+b}-a
\end{equation*}
How can I obtain some positive lower-bound?
 [/INST] Maybe..
$$
\sqrt{1+x} = 1 + \frac{1}{2} x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + O(x^4). 
$$
So, for small enough $x$,
$$
\sqrt{1+x} > 1 + \frac{1}{2}x - \frac{1}{8}x^2
$$
hence
$$
\sqrt{a^2 + b} = a \sqrt{1 + (b/a^2)} > a\Big(1 + \frac{b}{2a^2} -\frac{b^2}{8a^4} \Big)
$$
when $a$ is much larger than $b$. So, in this case,
$$
\sqrt{a^2+b} - a > \frac{b}{2a} -\frac{b^2}{8a^3}
$$
</s>