TITLE: Bennett's Inequality to Bernstein's Inequality
QUESTION [8 upvotes]: Bennett's Inequality is stated with a rather unintuitive function,
$$
h(u) = (1+u) \log(1+u) - u
$$
See here.  I have seen in multiple places that Bernstein's Inequality, while slightly weaker, can be obtained by bounding $h(u)$ from below,
$$
h(u) \ge \frac{ u^2 }{ 2 + \frac{2}{3} u}
$$
and plugging it back into Bennett's Inequality.  However, I can't see where this expression comes from.  Could someone point me in the right direction?

REPLY [2 votes]: Actually, the OP is asking where $\frac{u^2}{2 + \frac23 u} $ comes from.
The answer is: $f(u) = \frac{u^2}{2 + \frac23 u}$ is the Pade $(2,1)$ approximation of $h(u)$ at $u = 0$.
As regards the proof for the Pade $(2,1)$ approximation to be a lower bound, let
$$g(u) = \ln (1 + u) - \frac{1}{1 + u}\left(u + \frac{u^2}{2 + \frac23 u}\right).$$
We have
$$g'(u) = \frac{u^3}{(1 + u)^2(3 + u)^2}.$$
Note that $g'(u) < 0$ on $(-1, 0)$, $g'(u) > 0$ on $(0, \infty)$, and $g(0) = 0$.
Thus, $g(u) \ge 0$ on $(-1, \infty)$. We are done.