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?
 A: 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.
A: 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,
$$
\sqrt{a^2+b} - a > \frac{b}{2a} -\frac{b^2}{8a^3}
$$
A: 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 $0 \le b \le 8a^2$.
A: 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).
