<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] Factor out an $a^2$ from the radical to get $a\sqrt{1+\frac{b}{a^2}}-a=a\left(\sqrt{1+\frac{b}{a^2}}-1\right)$
Which can then be expanded for $\left|\frac{b}{a^2}\right|<1$, which is true for $a \gg b > 0$.
This expansion, to first order, is $a\left(1+\frac{b}{2a^2}-1\right)=\frac{b}{2a}$.
EDIT: Forgot that you were looking for meaningful bounds.
For a lower bound, you need to take the expansion to second order, so $a\left(1+\frac{b}{2a^2}-\frac{b^2}{8a^4}-1\right)=\frac{b}{2a}-\frac{b^2}{8a^3}$.
For an upper bound, you need the third order expansion, so $a\left(1+\frac{b}{2a^2}-\frac{b^2}{8a^4}+\frac{b^3}{16a^6}-1\right)=\frac{b}{2a}-\frac{b^2}{8a^3}+\frac{b^3}{16a^5}$
So overall, we have $\frac{b}{2a}-\frac{b^2}{8a^3}<\sqrt{a^2+b}-a<\frac{b}{2a}-\frac{b^2}{8a^3}+\frac{b^3}{16a^5}$
These inequalities are true for $0<\frac{b}{a^2}<1$, which is true in this case.
</s>