I'm trying to prove a bound of the form $\log(x + 1) \leq x – \frac{x^{2}}{a}$ for each $x \in (-1, f(a)]$. Here $a > 0$ is a fixed constant, and $f(a)$ is a function which depends on $a$. I'm trying to understand $f(a)$ better.
To be precise, for some fixed $a > 0$ one can consider the function $h_{a} \colon (-1, \infty) \to \mathbb{R}$, such that
$$h_{a}(x) := x – \frac{x^{2}}{a} – \log\left(x + 1\right)$$
By numerically plotting $h_{a}(x)$, for various values of $a \in \{1, 2, \ldots, 40\}$,
it is clear that $h_{a}(x) \geq 0$ for $x \in (-1, f(a)]$ and $h_{a}(x) < 0$, thereafter.
See plots below for different trajectories of $h_{a}(x)$ plotted using Sagemath. What is plotted is $h_{a}(x)$ for $a \in \{5, 10, 15, 20\}$.
$h_{a}(x)$ for $a \in \{5, 10, 15, 20\}$" />
Based on such plots, it is clear that $h_{a}(x) \geq 0$ for $x \in (-1, f(a)]$, where $f(a)$ is an increasing function in $a$. The $f(a)$ value is the one where $h_{a}(x)$ intersects the $x$ axis the second time and is negative thereafter.
My question is, how do I rigorously show that such an $f(a)$ value exists for each $a$ such that $h_{a}(x) \geq 0$ for $x \in (-1, f(a)]$? And how to prove that $f(a)$ is increasing in $a$? I would also like to understand how $f(a)$ behaves in $a$, e.g., useful bounds, or a good approximation thereof.
Update: Thanks to comments from @Arnaldo, @Gregory, I had reversed the required inequality as a typo. This is now corrected.
Best Answer
Some thoughts:
We split into three cases.
Case 1: $a = 2$
We have $\ln (x + 1) \le x - x^2/2$ for all $x \in (-1, 0]$, and $\ln (x + 1) > x - x^2/2$ for all $x > 0$.
Case 2: $a > 2$
Fact 1: For each fixed $a > 2$, the equation $\ln (x + 1) = x - x^2/a$ has exactly one positive real solution, denoted by $g(a)$.
Fact 2: For each fixed $a > 2$, we have $\ln(x + 1) \le x - x^2/a$ for all $x \in (-1, g(a)]$, and $\ln(x + 1) > x - x^2/a$ for all $x > g(a)$, where $g(a)$ is defined in Fact 1.
Case 3: $0 < a < 2$
Fact 3: For each fixed $0 < a < 2$, the equation $\ln (x + 1) = x - x^2/a$ has exactly one negative real solution, denoted by $h(a)$.
Fact 4: For each fixed $0 < a < 2$, we have $\ln(x + 1) \le x - x^2/a$ for all $x \in (-1, h(a)]\cup\{0\}$, and $\ln(x + 1) > x - x^2/a$ for all $x > h(a)$ and $x \ne 0$, where $h(a)$ is defined in Fact 3.
Proof of Fact 1:
Let $h_a(x) := x - x^2/a - \ln(x + 1)$. We have $h_a'(x) = 1 - 2x/a - \frac{1}{x + 1} = \frac{x(a - 2x - 2)}{a(x + 1)}$. Thus, $h_a'(x) > 0$ on $(0, a/2 - 1)$, and $h_a'(x) < 0$ on $(a/2 - 1, \infty)$, and $h_a'(a/2 - 1) = 0$.
Since $h_a(0) = 0$ and $h_a(\infty) = -\infty$, there exists exactly one $x_0 > 0$ such that $h_a(x_0) = 0$.
We are done.
Proof of Fact 2:
According to the proof of Fact 1, clearly, $h_a(x) \ge 0$ on $[0, g(a)]$, and $h_a(x) < 0$ on $x > g(a)$.
We have $h_a'(x) < 0$ on $(-1, 0)$. Also, $h_a(0) = 0$. Thus, $h_a(x) \ge 0$ on $(-1, 0)$.
We are done.