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?

The first three terms of $(1+x)^{\frac12}$ are $1 + \frac12 x - \frac18 x^2$. And you can check for yourself that $$\left(1 + \frac12 x - \frac18 x^2\right)^2 = 1 + x - \frac18 x^3 + \frac{1}{64}x^4$$
which is $\le 1+x$ whenever $\frac18 x^3 \ge \frac{1}{64}x^4$, i.e. for $0 \le x \le 8$.
Now just put $x = \frac{b}{a^2}$ to get
$$\sqrt{a^2+b} - a \ge \frac{b}{2a} - \frac{b^2}{8a^3}$$
whenever $0 \le b \le 8a^2$.
