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?

By the mean value theorem,
$$\sqrt{1 + x} - 1 = f(1 + x) - f(1) = x f'(c)$$
where $f$ is square root, $f'$ is its derivative,
and $c$ is some point in $[1, 1+x]$.
We need a lower bound and $f'$ is decreasing,
so $c$ is at worst $1 + x$ and we obtain
$$x f'(c) ≥ xf'(1 + x) = \frac{x}{2\sqrt{1 + x}}.$$
Backtrack: you want to underestimate $\sqrt{a^2 + b} - a$,
which is $a(\sqrt{1 + b/a^2} - 1)$
and we can let $x$ be $b/a^2$:
$$\sqrt{a^2 + b} - a ≥ a\frac{b/a^2}{2\sqrt{1 + b/a^2}} = \frac{b}{2\sqrt{a^2+b}}.$$
For an upper bound, $c$ is at best $1$ so
$$\sqrt{a^2 + b} - a ≤ \frac{b}{2a}$$
which is trivial (cf. completing the square).
