Does the following inequality hold true? I'm studying on a Jurimetrics problem where I want to design an optimal responsibility allocation mechanism between plaintiffs and defendants. But I'm right now facing the following problem, and I just have no idea how to prove it after several strives:
$$\frac{(1-x)^{2}-(1-y)^{2}}{1-x}>$$
$$\frac{(1-x-a)^{2}-(1-y-b)^{2}}{1-x-a}，$$ where $$x\in\left( 0,1 \right),y\in\left( 0,1 \right)$$, $$1>(x+a)>0, 1>(y+b)>0$$ and $1>a>b>0$ ?

 A: This inequality is false if e.g. $$x=\frac{5}{8},y=\frac{15}{16},a= \frac{7}{8},b=\frac{3}{4}.$$
(Then the left-hand side of your inequality is $\dfrac{140}{384}$ while its right-hand side is $\dfrac{171}{384}$.)

With the additional condition that $1 > x + a > 0$ and $1 > y + b > 0$, the inequality is true. According to Mathematica:


Here is a "human" proof of the inequality
with the additional condition that $1 > x + a > 0$ and $1 > y + b > 0$:
Letting $U:=1-x$, $V:=1-y$, $u:=U-a$, $v:=V-b$, we reduce the inequality in question to the inequality
$$U-\frac{V^2}U>u-\frac{v^2}u \tag{1}$$
if $U>u>0$, $V>v>0$, and $U>V-v+u$.
The left-hand side of (1) is increasing in $U$. So, (1) reduces to the inequality
$$d:=V-v+u-\frac{V^2}{V-v+u}-\Big(u-\frac{v^2}u\Big)\ge0. \tag{2}$$
But
$$ d=\frac{(u - v)^2 (V-v)}{(V-v+u)u}\ge0,$$
and we are done.
Remark: Here we only used the conditions $U>V-v+u$, $V>v$, and $u>0$, which are equivalent to the conditions $a>b$, $a>0$, and $1>x+a$ (respectively). None of the conditions $x>0$, $b>0$, $a<1$, $x+a>0$, $1 > y + b > 0$ was used.
