The inequalities are not equivalent. The inequality above holds if $a=3$ and $x=\frac{4}{3}$: $x + \frac{1}{2}x(1-x)a = \frac{4}{3} + \frac{1}{2}\left(\frac{4}{3}\right)\left(-\frac{1}{3}\right)(3) = \frac{4}{3} - \frac{2}{3} = \frac{2}{3},$ but this clearly does not satisfy $0\leq x \leq 1$, $|a|\lt 2$.
Here's how I found the counterexample.
If $a=0$, then the inequality is equivalent to $0\leq x \leq 1$.
If $a\neq 0$, let's divide the inequality into two quadratic inequalities: $0\leq x + \frac{1}{2}x(1-x)a \leq 1.$
Consider first $ 0 \leq x + \frac{1}{2}x(1-x)a = x\left(\left(1+\frac{a}{2}\right) - \frac{a}{2}x\right).$
The product is nonnegative if and only if we have both $x\leq 0$ and $1+\frac{a}{2} \leq \frac{a}{2}x$; or $x\geq 0$ and $1 + \frac{a}{2}\geq \frac{a}{2}x$.
If $x\geq 0$, then we need $1+\frac{a}{2}\geq \frac{a}{2}x$.
- If $a\gt 0$, then this is equivalent to $\frac{2}{a}+1\geq x$, so we need $0\leq x \leq 1+\frac{2}{a}$.
- If $a\lt 0$, then this is equivalent to $\frac{2}{a}+1\leq x$; if $-2\leq a \lt 0$, then we just need $x\geq 0$. If $a\leq -2$, then we need $1+\frac{2}{a}\leq x$.
If $x\leq 0$, then we need $1 + \frac{a}{2}\leq \frac{a}{2}x$.
- If $a\gt 0$, then this is equivalent to $\frac{2}{a}+1\leq x$, which is impossible if $x\leq 0$.
- If $a\lt 0$, then we need $\frac{2}{a}+1\geq x$. If $a\leq -2$, then $1+\frac{2}{a}\geq 0\geq x$, so this holds. If $-2\leq a\lt 0$, then we need $x\leq 1+\frac{2}{a}$.
In summary:
If $a\neq 0$, then the left inequality holds if and only if one of the following happens:
- $a\gt 0$ and $0\leq x \leq 1+\frac{2}{a}$; or
- $-2\leq a \lt 0$ and $x\geq 0$; or
- $-2\leq a\lt 0$ and $x\leq 1+\frac{2}{a}$; or
- $a\leq -2$ and $1+\frac{2}{a}\leq x$; or
- $a\leq -2$ and $x\leq 0$.
Now look at the right hand inequality $\left(1+\frac{a}{2}\right)x - \frac{a}{2}x^2 \leq 1$ which is equivalent to $ax^2 - (2+a)x + 2\geq 0.$ The two roots of this equation are $\frac{2+a+|a-2|}{2a}\quad\text{and}\quad \frac{2+a-|a-2|}{2a};$ Depending on the sign of $a-2$, one of the roots is $1$, the other root is $\frac{2}{a}$.
If $a\gt 0$, then the inequality is equivalent to not being between the two roots. If $a\lt 0$, then the inequality is equivalent to being between the two roots.
Thus:
Case 1. $a\leq -2$. We need either $x\leq 0$ and between $1$ and $\frac{2}{a}$; or $x\geq 1+\frac{2}{a}$ and between $1$ and $\frac{2}{a}$. So we must have either $\frac{2}{a}\leq x \leq 0$; or $1+\frac{2}{a}\leq x\leq 1$.
Case 2. $-2\leq a \lt 0$. Either $x\geq 0$ and between $\frac{2}{a}$ and $1$; or $x\leq 1 + \frac{2}{a}$ and between $1$ and $\frac{2}{a}$. That means, if $-2\leq a \lt 0$, we need either $0\leq x \leq 1$, or $\frac{2}{a}\leq x \leq 1+\frac{2}{a}$.
Case 3. $0\lt a\leq 2$. We need $0\leq x \leq 1+\frac{2}{a}$, and $x$ not between $1$ and $\frac{2}{a}$. This means we need $0\leq x \leq 1$ or $\frac{2}{a}\leq x \leq 1+\frac{2}{a}$.
Case 4. $2\leq a$. We need $0 \leq x \leq 1 + \frac{2}{a}$, and $x$ not between $\frac{2}{a}$ and $1$. This means we need $0\leq x \leq \frac{2}{a}$ or $1\leq x \leq 1 + \frac{2}{a}$. This is where the counterexample above came from.
In summary:
The inequality $0 \leq x + \frac{1}{2}x(1-x)a\leq 1$ holds if and only if:
- $a=0$ and $0\leq x \leq 1$; or
- $a\leq -2$ and $x\leq \frac{2}{a}$; or
- $a\leq -2$ and $1+\frac{2}{a}\leq x \leq 1$; or
- $-2\leq a \lt 0$ and $0\leq x \leq 1$; or
- $-2\leq a \lt 0$ and $\frac{2}{a}\leq x \leq 1+\frac{2}{a}$; or
- $0\lt a \leq 2$ and $0\leq x \leq 1$; or
- $0\lt a\leq 2$, and $\frac{2}{a}\leq x \leq 1+\frac{2}{a}$; or
- $2\leq a$ and $0\leq x\leq \frac{2}{a}$; or
- $2\leq a$ and $1\leq x \leq 1+\frac{2}{a}$.
And I hope I didn't mess up the cases... but the example above, at least, is correct.
From this, we can deduce pharmine's result; for the inequality to hold for all $x$ with $0\leq x \leq 1$, we need either $a=0$, $-2\leq a\lt 0$, or $0\lt a\leq 2$. In all other cases, some values on $[0,1]$ will not satisfy the inequality.