When exactly does this inequality hold/fail? Let $\alpha, \beta, x, y \in [0, 1]$. And suppose that $y + \beta \leq 1$, $x + y \leq 1$, $x + \alpha\leq 1$. Then, can anyone tell me when exactly does the following inequality fail/hold? 
$\alpha(1 - y- \beta) + \beta(1 - x) - xy \geq 0$
I've already figured out that $\beta \geq x$ is a sufficient condition for the inequality to hold. However, I've found both cases where it holds and where it fails for $\beta < x$. What are the missing conditions to characterise when the inequality holds/fails here?
 A: Denote the LHS of the inequality by $f$ where
$$f(x,y,\alpha,\beta)=\alpha(1-y)+\beta(1-x)-\alpha\beta-xy$$
Clearly $f$ is strictly decreasing in $x$ and $y$. When $x+y=1$, we have
$$f(x,1-x,\alpha,\beta)=(1-\alpha-x)(\beta-x).$$
Now, this quadratic has roots $\beta$ and $1-\alpha$, and so is nonnegative when $x\leq \min\{1-\alpha,\beta\}$ or when $x\geq \max\{1-\alpha ,\beta\}$. 
The working above shows that the following sets of conditions are sufficient for nonnegativity of $f$


*

*$\alpha,\beta\in[0,1]$ $x\in[0,\min\{1-\alpha,\beta\}]$, $y\geq 0$ $x+y\leq 1$

*$\alpha,\beta\in[0,1]$ $x\geq\max\{1-\alpha ,\beta\}$, $y\geq 0$ $x+y\leq 1$


Note that you do not need the condition $y\leq 1-\beta$ as a separate condition in either case. However for the second set of conditions, $y\leq 1-\beta$, is implied by $x\geq \beta$ and $x+y\leq 1$.
Since $f(x,y,\alpha,\beta)=f(y,x,\beta,\alpha)$, by symmetry the following conditions are also sufficient


*$\alpha,\beta\in[0,1]$ $x\geq 0$, $y\in[0,\min\{\alpha,1-\beta\}]$, $x+y\leq 1$

*$\alpha,\beta\in[0,1]$ $x\geq 0$, $y\geq\max\{\alpha,1-\beta\}$, $x+y\leq 1$

