Inequality - L'Hôpital's rule, logarithms Let's consider a function: (EDIT here: the number "2" was lacking)
$\mathcal{B}(x) = \frac{2}{x^2} \big( (1+x) log(1+x) - x \big)$
I need to prove inequality:
$ (1 + \frac{1}{3}x)\, \mathcal{B}(x) \geq 1 \, , \qquad x>0 \, ,$
using a hint: "Apply L'Hôpital's rule twice to reduce left-hand side to
$ \frac{1 + \frac{1}{3}x ^\ast}{1 + x^\ast} + \frac{2}{3} log (1 + x^\ast)$
for some $x^\ast$ less that $x$. Then use: $ \quad log(1+x^\ast) \geq \frac{x^\ast}{1 + x^\ast} \,$".
To be honest, I need a hint how to make use of this hint.
(EDIT:)
What I can do is compute: 
$f(x) =  (1 + \frac{1}{3}x) \big( (1+x) log(1+x) - x \big) \\
f'(x) = \frac{1}{3} \big( (1+x) log(1+x) - x \big) +  (1 + \frac{1}{3}x)log(1+x) \\
f''(x) = \frac{(1 + \frac{1}{3}x)}{1+x} + \frac{2}{3}\log(1+\frac{1}{3}) \\
g(x) = \frac{1}{2}x^2\\
g'(x) = x \\
g''(x) = 1\\
\underset{x \rightarrow 0^+}{\lim}f(x)=\underset{x \rightarrow 0^+}{\lim}f'(x)=\underset{x \rightarrow 0^+}{\lim}g(x)=\underset{x \rightarrow 0^+}{\lim}g'(x)=0 \\
\underset{x \rightarrow 0^+}{\lim}f''(x) = 1 = \underset{x \rightarrow 0^+}{\lim}g''(x)$
Since all of functions above are continuous and differentiable on $(0,\infty)$, limits of $f$, $f$, $g$, $g'$ when $x \rightarrow 0^+$ are equal to $0$ and limits of $f''$ and $g''$ are finite and $g''(x) \ne 0$, we obtain:
$$ \underset{x \rightarrow 0^+}{\lim}(1 + \frac{1}{3}x)\, \mathcal{B}(x) = 1 $$
But I don't understand the part of hint about "reducing the left hand side" to prove the inequality. Taking what I computed above and showing that the function is non-decreasing would be sufficient. But I want to understand method suggested by hint.

There is something strange in this problem, since, built at $x=0$, a Taylor expansion gives $$\mathcal{B}(x) = \frac{1}{x^2} \big( (1+x) \log(1+x) - x \big)=\frac{1}{2}-\frac{x}{6}+\frac{x^2}{12}-\frac{x^3}{20}+O\left(x^4\right)$$ Applying L'Hopital twice, you end with $$\frac{1}{2 (x+1)}$$ which just confirms the limit at $x=0$.
Then $$f(x)=(1 + \frac{x}{3})\, \mathcal{B}(x)=\frac{1}{2}+\frac{x^2}{36}-\frac{x^3}{45}+O\left(x^4\right)$$ Function $f(x)$ is continuously increasing and $f(x) >1$ only if $x>39.1242$.
