Prove $\log(x + 1) \leq x - \frac{x^{2}}{a}$ for fixed $a > 0$ 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.
 A: We have $h_a'(x)=1-2x/a-1/(x+1)$ which has the unique root $x^*=a/2-1$ over the positive reals.

Case 1: $a>2$. This case is so that $x^*>0$.
It is easy to show that $h_a''(x)<0$ for all $x>\sqrt{a/2}-1$ so we know that $h_a(x)$ is strictly decreasing in that interval.
Note that $x^*>\sqrt{a/2}-1$. Since $h_a(x^*)=(a^2-4)/4a-\log(a/2)>0$ and $h_a(a)=-\log(a+1)<0$, by the intermediate value theorem, $f(a)$ must exist and lies in the interval $(x^*,a)$; that is, we have the loose bounds $a/2-1<f(a)<a$.
A conjectured tight upper bound of $f(a)$ is $g(a)=a-\log(a+1)$. To prove this, it suffices that $$h_a(g(a))=\log(a+1)-\frac{\log^2(a+1)}a-\log(a+1-\log(a+1))<0$$ holds for all $a>0$. Let $b=\log(a+1)$, so that equivalently, $$b-\frac{b^2}{e^b-1}-\log(e^b-b)<0$$ for all $b>0$. I can't immediately think of a quick way to prove this.
Case 2: $0<a<2$. This case is so that $x^*<0$.
Here, we have $h_a'(x)<0$ for $-1<x<a/2-1$ and $h_a'(x)>0$ for $a/2-1<x<0$. By a similar argument, since $h_a(x^*)<0$ and $\lim\limits_{x\to-1^+}h_a(x)=+\infty$, we know that $f(a)$ must exist and lies in the interval $(-1,x^*)$; that is, we have the bounds $-1<f(a)<a/2-1$.

In both cases, to see why $f(a)$ must be increasing, note that for any $\varepsilon>0$ we have$$h_{a+\varepsilon}(x)=h_a(x)+\frac{\varepsilon x^2}{a(a+\varepsilon)}.$$ Since the remainder term is positive, the fact that $h_a(x)$ is strictly decreasing means it requires a larger value of $x$ such that $h_{a+\varepsilon}(x)=0$; that is, $f(a+\varepsilon)>f(a)$.
A: 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.
A: For the existence: For all $a$, we have $$\lim_{x \rightarrow -1^{+}} \left( \frac{x^2}{a}+\log(x+1)-x \right)=-\infty.$$
Hence there always exists a $f(a)$ such that $\frac{x^2}{a}+\log(x+1)-x <0$ for all $x \in (-1,f(a)]$.
For the exact best value of $f(a)$.
Clearly, it is sufficient to only consider the function $g(x)=\frac{x-\log(1+x)}{x^2}$. Then, we see that:
$$g'(x)=\frac{\log(1-\frac{x}{x+1})+\frac{x}{x+1}-\frac{x^2}{2(x+1)}}{-\frac{1}{2}x^3},$$
which means $g'(x)< 0$  on $(-1,0)$  and $g'(x)>(0,\infty)$.
Hence, the best value of $f(a)$ is $g^{-1}( \frac{1}{a})$ for $a \in (0,2)$  and $f(a)=\infty$ for all $a \ge 2$.
A: Want
$\log(x + 1) 
\leq x - \frac{x^{2}}{a}
$
for
$-1 < x < f(a)
$.
I first show that
$f(a)=2a-1$ works,
and then improve this to
$\dfrac32(2a-1)
$.
Note:
this is for $x > 0$.
$\ln(1+x)
=\int_0^x \dfrac{dt}{1+t}
$.
$\begin{array}\\
\ln(1+x)-x
&=\int_0^x \dfrac{dt}{1+t}-x\\
&=\int_0^x \left(\dfrac1{1+t}-1\right)dt\\
&=\int_0^x \dfrac{-t}{1+t}dt\\
x-\ln(1+x)
&=\int_0^x \dfrac{t}{1+t}dt\\
\end{array}
$
$\begin{array}\\
x-\ln(1+x)-\dfrac{x^2}{a}
&=\int_0^x \dfrac{t}{1+t}dt-\dfrac{x^2}{a}\\
&=\int_0^x \left(\dfrac{t}{1+t}-\dfrac{t}{2a}\right)dt\\
&=\int_0^x \dfrac{2at-t(1+t)}{2a(1+t)}dt\\
&=\int_0^x \dfrac{(2a-1)t-t^2}{2a(1+t)}dt\\
\end{array}
$
If
$(2a-1)t-t^2 \ge 0$
then
$x-\ln(1+x)-\dfrac{x^2}{a}
\ge 0$.
This is
$t \le 2a-1
$.
Therefore we can choose
$f(x)=2a-1$.
Also
$\begin{array}\\
x-\ln(1+x)-\dfrac{x^2}{a}
&=\int_0^x \dfrac{(2a-1)t-t^2}{2a(1+t)}dt\\
&>\int_0^x \dfrac{(2a-1)t-t^2}{2a(1+x)}dt\\
&=\dfrac1{2a(1+x)}\int_0^x ((2a-1)t-t^2)dt\\
&=\dfrac1{2a(1+x)}\left(\dfrac{(2a-1)x^2}{2}-\dfrac{x^3}{3}\right)\\
&=\dfrac1{2a(1+x)x^2}\left(\dfrac{(2a-1)}{2}-\dfrac{x}{3}\right)\\
&\ge 0
\qquad\text{when }x \le \dfrac32(2a-1)\end{array}
$
