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?

While the answer by @Arash outlines the general strategy of deriving the inequality,
the explicit computation is missing.
I am providing it here in the hope that it will be helpful for future readers.
The actual computation of the derivative is highly simplified by first simplifying
the last term in the definition of
$$
  f(x)
  = (1 + x) \ln(1 + x) - x - \frac{x^2}{2 + \frac{2}{3} x}
$$
as follows:
$$
  \frac{x^2}{2 + \frac{2}{3} x}
  = \frac{3}{2} \frac{x^2}{3 + x}
  = \frac{3}{2} \frac{x \cdot (3 + x) - 3 x}{3 + x}
  = \frac{3}{2} x - \frac{9}{2} \frac{x}{x + 3}.
$$
Therefore, we see
\begin{align*}
  \require{cancel}
  f(x)
  & = (1 + x) \ln (1 + x) - \frac{5}{2} x + \frac{9}{2} \frac{x}{x + 3} , \\
  f'(x)
  & = \ln(1 + x) + 1 - \frac{5}{2} + \frac{9}{2} \frac{x+3 - x}{(x+3)^2} \\
  & = \ln(1 + x) - \frac{3}{2} + \frac{3^3}{2} (x+3)^{-2}, \\
  f''(x)
  & = \frac{1}{1 + x} - 3^3 \cdot (x+3)^{-3} \\
  & = \frac{(x+3)^3 - 27 \cdot (1 + x)}{(1 + x) (3+x)^3} \\
  & = \frac{x^{3} + 9 x^{2} + \bcancel{27 x} + \cancel{27} - \cancel{27} - \bcancel{27 x}}
           {(1 + x) (3+x)^3} \\
  & = \frac{x^{3} + 9 x^{2}}
           {(1 + x) (3+x)^3}
    \geq 0
\end{align*}
for $x \geq 0$, which is the only case we are interested in.
Since $f(0) = 0$ and $f'(0) = 0$, this implies $f' \geq 0$ and then $f \geq 0$
on $[0,\infty)$.
