Better proof of inequality $x - (1 + x) \log(1+x) \leq -\frac{x^2}{2(1+x)}$ for $x > 0$ The following inequality is valid for all positive real $x$,
$$
x - (1+x)\log(1+x) \leq \frac{-x^2}{2(1+x)}.
$$
It is possible to show that this is true by considering the function
$$
f(x) := x - (1+x)\log(1+x)+ \frac{x^2}{2(1+x)}.
$$
By differentiation it is possible to check that $f(x)$ attains it maximum on the nonnegative reals at $x = 0$.
However, is there a cleaner, more obvious way to see that this is true? Specifically, I would like a solution that does not require me to analyze the monotonicity of $f$ via differentiation if possible. I tried Taylor expansion, but do not get this inequality.
 A: Better proof:
$\log(1+y)\le y-\dfrac{1}{2}y^2\quad$ for all $\;y\in\left]-1,0\right]\;.\quad\color{blue}{(*)}$
For all $\;x\ge0\;,\;$ it results that $\;y=\dfrac{1}{1+x}-1\in\left]-1,0\right],$ hence, by applying $(*)$, we get that
$\log\left(1+\dfrac{1}{1+x}-1\right)\le\dfrac{1}{1+x}-1-\dfrac{1}{2}\left(\dfrac{1}{1+x}-1\right)^2,$
$-\log\left(1+x\right)\le\dfrac{1}{1+x}-1-\dfrac{1}{2}\dfrac{x^2}{(1+x)^2}\;,$
$-(1+x)\log\left(1+x\right)\le 1-(1+x)-\dfrac{x^2}{2(1+x)}\;,$
$-(1+x)\log\left(1+x\right)\le -x-\dfrac{x^2}{2(1+x)}\;,$
$x-(1+x)\log(1+x)+\dfrac{x^2}{2(1+x)}\le0\;,\;$ for all $\;x\ge0\;.$
A: Yet another way to obtain such a bound is as follows. Set $\phi(\delta) = (1+\delta)\log(1+\delta)$.
One then checks that
$$
\phi(0) = 0, \quad \dot{\phi}(0) = 1, \quad \mbox{and} \quad \ddot{\phi}(z) = \frac{1}{1+z}.
$$
Therefore, by Taylor's theorem, for some $\tilde x \in (0, x)$:
$$
x - (1+x)\log(1+x) = x - \phi(x) = x - \Big\{\phi(0) + \dot{\phi}(0) x + \ddot{\phi}(\tilde x) \frac{x^2}{2}\Big\} \leq \frac{-x^2}{2(1+x)}
$$
as claimed. (The final inequality uses $\ddot{\phi}(\tilde x) \geq \tfrac{1}{1 + x}$.)
