Determining $a$ and $b$ such that the expression is negative?

Question

For what values of $a$ and $b$,

Is $2a(x-1)(xy-x^2)+2b(y-1)(y-xy+2-2y^2) < 0$ for $x,y \in \Bbb{R}$ except $x=1$ and $y=1$.

I am not seeing a method to find it, any clever observation?

I tried making a complete square but could not?

Any help?

Answer 1

Let $x=y$, the first term vanishes.

$$2b(y-1)(-3y^2+y+2) <0$$

$$2b(y-1)(-3y-2)(y-1) <0$$

Now let $-3y-2=0$

and we get a contradiction since at $x=y=-\frac23$, we get $0<0$.

Hence no such $a,b$ exists.

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