two variable nonhomogeneous inequality Let $$x\ge 0,y\ge 0,x\neq 1,y \neq 1$$Prove the inequality
$$\dfrac {x}{(y-1)^2} +\dfrac {y}{(x-1)^2} \ge \dfrac {x+y-1}{(x-1)(y-1)} $$
 A: hint: The endpoints case you can handle with ease, for more general case that: $x, y > 1\to x(x-1)^2 +y(y-1)^2 \geq (x+y-1)(xy-(x+y-1))\iff x(x^2-2x+1)+y(y^2-2y+1)\geq xy(x+y-1)-(x+y-1)^2\iff (x^3+y^3)-2(x^2+y^2)+(x+y)\geq xy(x+y-1)-(x^2+y^2+1+2xy-2x-2y)\iff f(x,y)=x^3+y^3-(x^2+y^2)+1+3xy-(x+y) -xy(x+y)\geq 0$. Taking partial derivatives:
$f_x=3x^2-2x+3y-1-2xy-y^2 = 0=f_y=3y^2-2y+3x-1-x^2-2xy\to f_x-f_y = 0\to (x-y)(4(x+y)-5)=0\to x=y , x+y = \dfrac{5}{4}$. Consider each case separately will yield desire result.
Note: If you want to avoid using calculus, there is another way ( I let you test it to see if it works) that you can prove this at this point:
You prove : $f(x,y) \geq f(y,y)$, and $f(y,y) \geq 0$. The latter inequality is $(y-1)^2 \geq 0$.
Another approach is let $a = x-1,b = y-1$, and assume $a,b > 0\to \dfrac{a+1}{b^2}+\dfrac{b+1}{a^2}\geq \dfrac{a+b+1}{ab}$.But this is quite simple...because $LHS =\dfrac{a}{b^2}+\dfrac{b}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\geq \dfrac{a^3+b^3}{a^2b^2}+\dfrac{2}{ab}\geq \dfrac{ab(a+b)}{a^2b^2}+\dfrac{2}{ab}=\dfrac{a+b+2}{ab} > RHS$
A: 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
$$\hat f(u,v)\geq(1-u)^2-u^2(5-8u)=(1-2u)^2(1+2u)\geq0\ .$$
