Show that $(1+u) \log (1+u) - u \ge \frac{u^2}{2(1+u/3)} $ This is used to go from Bennett's inequality to Bernstein's inequality.
Yet I don't understand how to prove the inequality.
Assume that $u > 0$, define
$$
h(u) = (1+u) \log (1+u) - u
$$
show that
$$
h(u) \ge \frac{u^2}{2(1+u/3)}
$$

My research :
Decomposing the function $h$ as a power series show that it is equivalent to
$$
\sum_{n=1}^\infty (-1)^n \frac{u^n}{n+1} \frac{n-1}{n (n+2)} \ge 0
$$
Sadly, I see no reason for this series to be positive.

SOLUTION
The twice differenciation technique given below works. However, I don't agree with the calculus, only on small things that don't change the result. Define
$$
g(u) = (1+u)\log (1+u) - u - \frac{3u^2}{2(3+u)}
$$
$$
g'(u) = \log (1+u) + 1 - 1 - \frac{3}{2} \frac{2u (3+u) - u^2}{(3+u)^2}\\
$$
Simplifying,
$$
g'(u) = \log (1+u) - \frac{3}{2} \frac{u(u+6 )}{(3+u)^2}  
$$
$$
g''(u) = \frac{1}{1+u} - \frac{3}{2} \frac{(2u +6)(3+u)^2 - 
2 (u+3)(u^2 + 6u) }{(3+u)^4}  
$$
Simplifying,
$$
g''(u) = \frac{1}{1+u} - \frac{3}{2} \frac{2u^2 +6u + 6u + 18 - 
2u^2 - 12u }{(3+u)^3}   =
 \frac{1}{1+u} - \frac{ 27  }{(3+u)^3} 
$$
Simplifying again,
$$
g''(u)= \frac{ u^3 + 3\times 3u^2 + 3 \times 9u + 27 - 27(1+u)}{(1+u)(3+u)^3} 
= \frac{ u^2(u+9)}{(1+u)(3+u)^3} > 0
$$
And the result is given by the reasoning of Clement.
 A: Typically, the brute-force, not-so-clever, usually conclusive method to prove $f\geq g$ is to study either $\frac{f}{g}$ or $f-g$ on the desired domain, and prove (by differentiation, invoking monotonicity, and evaluation at specific points) that either the ratio is at least 1, or the difference at least 0. In this case, the difference seems nicer, as we want to  "get rid of the logarithm" by differentiation as quickly as we can, and that's not going to happen by differentiating the ratio.
Let $g$ be defined by $g(u) = h(u)-\frac{u^2}{2(1+\frac{u}{3})}$ for $u>0$.
$g$ is then smooth on its domain, and we have
$$\begin{align}
g'(u) &= \ln(1+u) - \frac{3u(u+6)}{2(u+3)^2}\\
g''(u) &=  \frac{u^2(u+9)}{(u+1)(u+3)^3} >0
\end{align}$$
for $u>0$. In particular, $g'$ is increasing, and since $\lim_{u\to 0^+} g'(u) = 0$, we know that $g'>0$.
So $g$ is increasing. But as $\lim_{u\to 0^+} g(u) = 0$ as well, this means $g>0$ on its domain; hence
$$
h(u) > \frac{u^2}{2(1+\frac{u}{3})} \qquad \forall u>0.
$$
A: Your inequality is related to the Padé approximation $$\frac{(x-1)(x+5)}{4x+2}$$ of $\log(x)$. In fact one can show (e.g. see below) that $$\log(x) \leq \frac{(x-1)(x+5)}{4x+2}$$ for all $x>0$. (This is a useful inequality to remember.) Substituting $x\leftarrow 1/x$ reverses the inequality: $$\log(x)= -\log\left(\frac1x\right) \geq -\frac{(\frac1x-1)(\frac1x+5)}{\frac4x+2} = \frac{(x-1)(5x + 1)}{x(2x+4)}.$$ As a direct consequence $$ x \log(x) - (x-1) \geq \frac{3(x-1)^2}{6+2(x-1)}.$$
Here is one way to prove the inequality for the Padé approximation. Fix  $x \geq 1$ and let $f(t) = (t-1)^2(x-t)$. Then $f(1)=f(x)=0$ and $f(t) \geq 0$ for $t \in[1, x]$. Integration by parts shows $$\int_1^x \frac{f'(t)}t \mathrm{d}t = \int_1^x \frac{f(t)}{t^2}\mathrm{d}t \geq 0.$$ Now the integral on the left can be easily computed: $$\int_1^x \frac{f'(t)}t \mathrm{d}t = (x-1)(x+5)/2-(2x+1)\log(x).$$ A similar argument works for $0<x\leq 1$ (now $f(t)\leq 0$ on the interval $[x,1]$).
