Bennett's Inequality to Bernstein's Inequality 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?

Here is how the approximation
could be derived.
$\begin{align}
h(u)
&=(1+u)\log(1+u)−u\\
&=(1+u)\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}−u\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}
+u\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}−u\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}
+\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^{n+1}}{n}−u\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}
+\sum_{n=2}^{\infty} \dfrac{(-1)^{n}u^{n}}{n-1}−u\\
&=u+\sum_{n=2}^{\infty} \dfrac{(-1)^{n-1}u^n}{n}
+\sum_{n=2}^{\infty} \dfrac{(-1)^{n}u^{n}}{n-1}−u\\
&=\sum_{n=2}^{\infty} u^n\left(\dfrac{(-1)^{n-1}}{n}
+\dfrac{(-1)^{n}}{n-1}\right)\\
&=\sum_{n=2}^{\infty} (-1)^{n}u^n\left(\dfrac{-1}{n}
+\dfrac{1}{n-1}\right)\\
&=\sum_{n=2}^{\infty} (-1)^{n}u^n\left(\dfrac{1}{(n-1)n}\right)\\
&=\sum_{n=2}^{\infty} \dfrac{(-1)^{n}u^n}{(n-1)n}\\
&= \dfrac{u^2}{2}-\dfrac{u^3}{6}+\dfrac{u^4}{12}
-\dfrac{u^5}{20}\pm ...\\
\end{align}
$
If we just look at the first two terms,
$\dfrac{u^2}{2}-\dfrac{u^3}{6}
=\dfrac{u^2}{2}(1-\dfrac{u}{3})
$,
and since
$\dfrac1{1+z}
=1-z+z^2 \pm ...
$,
$1-\dfrac{u}{3}
\sim \dfrac1{1+u/3}
$.
Therefore
$h(u)
\sim \dfrac{u^2}{2}\dfrac1{1+u/3}
= \dfrac{u^2}{2+2u/3}
$.
This shows how the approximation
could be derived.
The next step is to see
how accurate the approximation is.
Let
$h^*(u) = \dfrac{u^2}{2}\dfrac1{1+u/3} = \dfrac{u^2}{2+2u/3} $.
Expanding the approximation,
$\begin{align}
h^*(u)
&\sim \dfrac{u^2}{2}\dfrac1{1+u/3}\\
&=\dfrac{u^2}{2}(1-\dfrac{u}{3}+\dfrac{u^2}{9}
-\dfrac{u^3}{27}\pm ...)\\
&=\dfrac{u^2}{2}-\dfrac{u^3}{6}+\dfrac{u^4}{18} -\dfrac{u^5}{54}\pm ... \\
\end{align}
$
Therefore
$\begin{align}
h(u)-h^*(u)
&=(\dfrac{u^4}{12} -\dfrac{u^5}{20}\pm ...)
-(\dfrac{u^4}{18} -\dfrac{u^5}{54}\pm ...)\\
&=u^4(\dfrac{1}{12}-\dfrac{1}{18}) -u^5(\dfrac{1}{20}-\dfrac{1}{54})\pm ...\\
&=\dfrac{u^4}{36} -\dfrac{17u^5}{540}\pm ... \\ 
\end{align}
$
It certainly looks like
$h(u) > h^*(u)$.
Once you have this conjecture,
you can try to prove it.
In this case,
a proof can be developed
by looking at the successive terms
in the difference
and showing that the terms are decreasing
and alternating in sign.
Note that if we let
$h^{**}(u)
= \dfrac{u^2}{2}-\dfrac{u^3}{6}
$,
$h(u)-h^{**}(u)
=\dfrac{u^4}{12} -\dfrac{u^5}{20}\pm ...
$,
and this appears to have an error
about three times as large.
