two variable nonhomogeneous inequality Let $$x\ge 0,y\ge 0,x\neq 1,y \neq 1$$Prove the inequality
$$\dfrac {x}{(y-1)^2} +\dfrac {y}{(x-1)^2} \ge \dfrac {x+y-1}{(x-1)(y-1)} $$

When $0\leq x<1<y$ or $0\leq y<1<x$ then the right side of the stated inequality is negative.
For the remaining cases we clear denominators and consider the function
$$f(x,y):=x(x-1)^2+y(y-1)^2-(x-1)(y-1)(x+y-1)\qquad(x\geq0, \ y\geq0)\ .$$
When $x>1$ and $y>1$ we write $x:=1+u$, $y:=1+v$ with $u>0$, $v>0$ and obtain after expanding
$$\hat f(u,v)=u^2-uv+v^2+(u+v)(u-v)^2\ .$$
This is $>0$ since the quadratic form $u^2-uv+v^2$ is positive definite.
When $0\leq x\leq y\leq 1$ (and similarly, when $0\leq x\leq y\leq1$) we denote the arithmetic mean of the two by $u$ and write
$$x:=u-v,\qquad y:=u+v$$
with $0\leq u\leq 1$ and $0\leq v\leq\min\{u,1-u\}$. In this way we obtain after expanding
$$\hat f(u,v)=(1-u)^2+v^2(8u-5)\ .$$
When $u\geq{5\over8}$ then obviously $\hat f(u,v)\geq0$. For ${1\over2}\leq u\leq{5\over 8}$ one has
$$\hat f(u,v)=(1-u)^2-v^2(5-8u)\geq (1-u)^2\bigl(1-(5-8u)\bigr)\geq0\ ,$$
since $8u-4\geq0$. Finally $0\leq u\leq{1\over2}$ leads to
$$\hat f(u,v)\geq(1-u)^2-u^2(5-8u)=(1-2u)^2(1+2u)\geq0\ .$$
