"Elegant" proof that $f(x) \geq g(x)$ Fix $\alpha\in(0,1)$, $\beta >1$ such that $\frac{\alpha\beta}{1-\alpha} \leq 1$; and define $f,g\colon [0,1]\to\mathbb{R}$ by
$$
f(x) = x\left(1-\left(\frac{1-x}{1-\alpha}\right)^\beta\right), \qquad g(x) = \frac{\alpha\beta}{1-\alpha}(x-\alpha) + \mathbf{1}_{[0,\alpha]}(x)\cdot \frac{\beta}{2} (x-\alpha)^2
$$
Note that $g$ is nondecreasing, that $f(0)=f(\alpha)=0=g(\alpha)$, $f(1)=1$, $g'(\alpha)=f'(\alpha) = \frac{\alpha\beta}{1-\alpha}$. I have verified for various settings of $\alpha,\beta$ satisfying the conditions above that the relation
$$
f \geq g
$$
holds on $[0,1]$; I also think that, if I really need to, I could prove it in a very ugly way, possibly involving an ungodly amount of differentiation.

Is there a "clean" and elegant way to prove that $f\geq g$?

(In the (unlikely/surprising to me) event that the condition $\frac{\alpha\beta}{1-\alpha} \leq 1$ does not suffice, adding $\alpha \leq 1/2$, or any $\alpha\beta \leq c$ for some absolute constant $c>0$ is still alright by me.)
Edit: I think that the $\frac{\beta}{2}$ coefficient can be replaced by $\frac{\beta}{2(1-\alpha)}$. The stronger inequality should still hold, not sure if that's harder to prove or if it strangely makes things a bit easier.
(The choice of that specific coefficient for the quadratic term is to ensure that $g'(0) \geq 0$, which I need somewhere else.)
 A: We will substitute
$$ s = \frac{x-\alpha}{1-\alpha} \qquad \text{and} \qquad \gamma = \frac{\alpha}{1-\alpha}. $$
Then we know that $s \leq 1$ and $\beta \gamma \leq 1$. Now using the inequality $0 \leq 1 - s \leq e^{-s}$, we get
\begin{align*}
\frac{f(x) - g(x)}{1-\alpha}
&= (s + \gamma)(1 - (1 - s)^{\beta}) - \beta \gamma s - \frac{\beta s^2}{2(1+\gamma)} \mathbf{1}_{\{s < 0\}} \\
&\geq (s + \gamma)(1 - e^{-\beta s}) - \beta \gamma s - \frac{\beta s^2}{2(1+\gamma)} \mathbf{1}_{\{s < 0\}}
\end{align*}
Write $h(s)$ for the last line. We show that $h(s) \geq 0$.

*

*First, we note that $h(0) = 0$.


*Differentiating $h(s)$ with respect to $s$,
$$ h'(s) = (1 - \beta \gamma) (1 - e^{-\beta s}) + \beta s e^{-\beta s} - \frac{\beta s}{1+\gamma} \mathbf{1}_{\{s < 0\}}. $$
From this, we know that $h'(s) \geq 0$ for $s > 0$ and hence $h(s) \geq 0$ for $s \geq 0$.


*Now we focus on the region $ s < 0$. Differentiating again,
\begin{align*}
h''(s)
&= \beta \left( e^{-\beta s} (2 - \beta s -\beta \gamma) - \frac{1}{1+\gamma} \right) \\
&\geq \beta \left( e^{-\beta s}- \frac{1}{1+\gamma} \right) \\
&\geq 0.
\end{align*}
This shows that $h$ is convex on $s \leq 0$. Moreover, since $h(0) = h'(0) = 0$, it follows that $h$ attains the minimum at $s = 0$ and hence $h(s) \geq 0$ on this region.
A: Remarks: We give a stronger result.
Fact 1: If $\alpha \in (0, 1)$ and $0 < \beta \le \frac{1}{\alpha}$
and $x \in [0, 1]$, then
$$x\left[1-\left(\frac{1-x}{1-\alpha}\right)^\beta\right]
\ge \frac{\alpha \beta}{1 - \alpha}(x - \alpha)
+ 1_{[0, \alpha]}(x)\cdot \frac{\beta}{2(1 - \alpha)}(x - \alpha)^2.$$

Proof of Fact 1:
When $x = 0, \alpha, 1$, clearly the desired inequality is true.
In the following, assume that $0 < x < 1$ and $x \ne \alpha$.
The desired inequality is written as
$$x\cdot \frac{1}{\beta}\left[1-\left(\frac{1-x}{1-\alpha}\right)^\beta\right]
\ge \frac{\alpha }{1 - \alpha}(x - \alpha)
+ 1_{[0, \alpha]}(x)\cdot \frac{1}{2(1 - \alpha)}(x - \alpha)^2.$$
Fact 2: Let $c > 0$.
Then $h(r) = \frac{1 - c^r}{r}$ is non-increasing on $(0, \infty)$.
(Proof: We have $h'(r) = - \frac{c^r}{r^2}[c^{-r} - 1 - \ln (c^{-r})] \le 0$ where we have used $u - 1 \ge \ln u$ for all $u > 0$. We are done.)
By Fact 2, it suffices to prove that
$$x\cdot \frac{1}{1/\alpha}\left[1-\left(\frac{1-x}{1-\alpha}\right)^{1/\alpha}\right]
\ge \frac{\alpha }{1 - \alpha}(x - \alpha)
+ 1_{[0, \alpha]}(x)\cdot \frac{1}{2(1 - \alpha)}(x - \alpha)^2.$$
We split into two cases:
Case I $~ 0 < x < \alpha$:
It suffices to prove that
$$\alpha x - \frac{\alpha }{1 - \alpha}(x - \alpha)
- \frac{1}{2(1 - \alpha)}(x - \alpha)^2 \ge \alpha x \left(\frac{1-x}{1-\alpha}\right)^{1/\alpha}$$
or
$$\ln \left(\alpha x - \frac{\alpha }{1 - \alpha}(x - \alpha)
- \frac{1}{2(1 - \alpha)}(x - \alpha)^2\right) \ge \ln \alpha + \ln x
+ \frac{1}{\alpha}\ln \frac{1-x}{1-\alpha}.$$
Let $F(x) = \mathrm{LHS} - \mathrm{RHS}$.
We have
$$F'(x) = - \frac{(\alpha - x)^2[\alpha + (1 - \alpha)x]}{(1 - x)\alpha x [\alpha^2 - x^2 + 2\alpha(1 - \alpha)x]} \le 0.$$
Also, $F(\alpha) = 0$.
Thus, $F(x) \ge 0$ for all $x\in (0, \alpha)$.
Case II $~ \alpha < x < 1$:
It suffices to prove that
$$\alpha x - \frac{\alpha }{1 - \alpha}(x - \alpha)
\ge \alpha x \left(\frac{1-x}{1-\alpha}\right)^{1/\alpha}$$
or
$$\ln \left(\alpha x - \frac{\alpha }{1 - \alpha}(x - \alpha)
\right) \ge \ln \alpha + \ln x
+ \frac{1}{\alpha}\ln \frac{1-x}{1-\alpha}.$$
The remaining is similar to the proof of Case I.
We are done.
