Prove that - $$3(x^2-x+1)(y^2-y+1) \geqq 2(x^2y^2 -xy+1)$$
My Approach I tried to solve this inequality using the A.M. - G.M. but I wasn't able to get the solution using it. I would appreciate if someone shows me the solution using the A.M. - G.M. if it is possible to do so
On
What follows couldn't have existed without the solution by S. Dolan and somewhat paraphrase it, because it comes from the "canonical factorization" of his quadratic. As it is too long for a comment, I write it as an answer.
Let $$f(x,y):=3(x^2-x+1)(y^2-y+1)-2(x^2y^2 -xy+1)\tag{1}$$
We have to prove that, for all $x,y$,
$$f(x,y) \geq 0. \tag{2}$$
It is consequence of the (non-evident!!) identity
$$4(y^2 - 3y + 3)f(x,y)=\underbrace{(2xy^2- 6xy - 3y^2+ 6x + 5y- 3)^2+3(y^2-3y+1)^2}_{\text{sum of squares}}\tag{3}$$
Indeed, quadratic $y^2-3y+3>0$, having a negative discriminant, is always $>0$, proving assertion (2).
I am fully aware that finding a factorization such as (3) directly is not evident. But maybe some future reader will be able to say that it can be found by a direct reasoning...
Remark : there is a "sister identity" of (3) obtained by exchanging the roles of $x$ and $y$... Could it be a source of a better understanding ?
I think the best way to solve this is to use the quadratic formula. Collecting terms as a quadratic in $x$ gives the inequality as $$(y^2-3y+3)x^2+(-3y^2+5y-3)x+3y^2-3y+1\ge 0.$$
Now $y^2-3y+3=(y-\frac{3}{2})^2+\frac{3}{4}>0$ and so this quadratic in $x$ is certainly positive for large $x$. The issue is whether it can ever cross the $x$-axis.
Let $a=y^2-3y+3,b=-3y^2+5y-3,c=3y^2-3y+1$. Then $$4ac-b^2=3(y^4-6y^3+11y^2-6y+1)=3(y^2-3y+1)^2\ge 0.$$
So, for values of $y$ such that $y^2-3y+1\ne 0,$ the quadratic in $x$ is always positive. For $y^2-3y+1= 0,$ the quadratic in $x$ touches the $x$-axis but does not cross it.
It might be of interest that equality occurs when $$x=y=\frac{3\pm \sqrt 5}{2}.$$