Show that $(1+x)\log(1+x)-x \geq x^2/4$ on $(-1, 4]$

Question

In Bolley, Villani [1] p. 340 the following "elementary inequality" is used for $-1 < x \leq 4$: $$ h(x) := (1+x)\log(1+x)-x \geq x^2/4, $$ "a consequence of the fact that $h(x)/x$ is nondecreasing".

I am struggling to see how this is a consequence.

Plotting shows that similar inequalities $h(x) \geq x^2/c$ hold on different domains for $c \geq 2$. The inequality cannot hold for $c < 2$ due to the asymptotics at $0$.

Since $\log(1+x) < x$ we also have $h(x) < x^2$ so that $x^2/4 \leq h(x) < x^2$. And $h$ is of course the integral of $\log(1+x)$.

Any suggestions on how to derive this inequality (especially from the hint) would be much appreciated.

[1] Bolley, Villani - Weighted Csiszár-Kullback-Pinsker inequalities and applications to transportation inequalities, https://afst.centre-mersenne.org/article/AFST_2005_6_14_3_331_0.pdf

Update: For $a > 0$ and $g(x)$ such that $g(a)/a = 1$ and $g(x)/x$ increasing, we have for $x \in [a, 4]$ that $$ g(x) = \left(\frac{g(x)}{x}\right) x \geq \left(\frac{g(a)}{a}\right) x = x $$ and $$ 4g(x) - x^2 \geq 4x - x^2 = x(4-x) \geq 0. $$ However here, even though $h(x) = x$ at $x = 0$, we have $h(x)/x = 0$ at $0$ so the reasoning above does not hold in this case.

Maybe this trick together with some informal reasoning was the idea behind the hint, but who knows what the authors had in mind... I have accepted xpaul's answer, which does not use the hint, for now.

Answer 1

Let $$ f(x)=(x+1)\ln(1+x)-x-\frac{x^2}{4}. $$ We have to show $f(x)\ge0$ in $(-1,4]$. Clearly $\lim_{x\to-1^+}f(x)=\frac34$. We can assume $f(-1)=\frac34$ and hence $f(x)$ is continuous in $[-1,4]$ and diffferentiable in $(-1,4)$. Note $$ f'(x)=-\frac{x}{2}+\ln(1+x), f''(x)=\frac{1-x}{2(1+x)} $$ and $x=1$ is the only point such that $f''(x)=0$. Therefore $f'(x)=0$ has two solutions $x=0$ and $x=c$, $c\in(1,4)$. Clearly $f''(0)=\frac12>0, f''(c)<0$ and hence $f(0)=0$ is a local minimum and $f(c)$ is a local maxim. Also note $f(-1)=\frac34>f(0), f(4)=5\ln5-8>f(0)$. Thus $f(x)$ reaches the global minimum at $x=0$ in $[-1,4]$, namely $f(x)\ge0$.

Answer 2

I'm as baffled as you are regarding the hint, but in order to show the inequality, it's enough to define and differentiate

$$\begin{align} f(x)&=(1+x)\log(1+x)-x-{x^2\over4}\\ f'(x)&=\log(1+x)-{x\over2}\\ f''(x)&={1-x\over2(1+x)} \end{align}$$

and note that $f(0)=f'(0)=0$ and $f(4)=5\log(5)-8\approx0.47\gt0$, with $f''(x)\gt0$ for $x\lt1$ and $f''(x)\lt0$ for $x\gt1$. This is enough to show that $f(x)\ge0$ for all $x\in(-1,4]$.