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.

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 $[x,1]$).
