An inequality involving three reals greater than or equal to 1 I have been trying to verify the inequality for $x\geq k, y\geq k,$ and $k\geq 1,$
$\left(\frac{\left(x-k\right)}{x+1}\right)\left(\frac{\left(y-k\right)}{y+1}\right)+\left(\frac{\left(x-k\right)}{x+1}\right)+\left(\frac{\left(y-k\right)}{y+1}\right)\geq\left(\frac{\left(xy-k^2\right)}{xy+1}\right) . $  Let me show my attempt.
Let us try to prove $\left(\frac{\left(x-k\right)}{x+1}\right)\left(\frac{\left(y-k\right)}{y+1}\right)+\left(\frac{\left(x-k\right)}{x+1}\right)+\left(\frac{\left(y-k\right)}{y+1}\right)-\left(\frac{\left(xy-k^2\right)}{xy+1}\right)\geq 0  $
But the left hand side of the above inequality is equal to
$$\frac{\left(x-k\right)\left(y-k\right)\left(xy+1\right)+\left(x-k\right)\left(y+1\right)\left(xy+1\right)+\left(y-k\right)\left(x+1\right)\left(xy+1\right)-\left(xy-k^2\right)\left(x+1\right)\left(y+1\right)}{\left(x+1\right)\left(y+1\right)\left(xy+1\right)}$$
$$=\frac{2x^2y^2-2kx^2y+k^2x-2kx-2kxy^2+2xy+2k^2xy-2kxy+x+2k^2-2k-2ky+y+k^2y}{\left(x+1\right)\left(y+1\right)\left(xy+1\right)}. $$
Is there anyway to conclude
$2x^2y^2-2kx^2y+k^2x-2kx-2kxy^2+2xy+2k^2xy-2kxy+x+2k^2-2k-2ky+y+k^2y\geq 0$ using factorization or any other technique?
One can observe that whenever $k=1,$
$$2x^2y^2-2kx^2y+k^2x-2kx-2kxy^2+2xy+2k^2xy-2kxy+x+2k^2-2k-2ky+y+k^2y=2xy(x-1)(y-1)\geq 0$$ and the result is true whenever $k=1.$
 A: I shall not mention the obvious symmetry.
What I did is to look at the minimum value of function
$$f(x,y)=\frac{x-k}{x+1}\times\frac{y-k}{y+1}+\frac{x-k}{x+1}+\frac{y-k}{y+1}-\frac{x y-k^2}{x y+1}$$
$$f(k,y)=\frac{(k-1) (k-y)}{(y+1) (k y+1)}$$ Its derivative cancels at
$$y_*=k+\frac{\sqrt{k^4+k^3+k^2+k}}{k}$$ and the second derivative test shows that this is a minimum.
Now, computing $$f(k,y_*)=\frac{2 k^2+k+1-2 \sqrt{k (k+1) \left(k^2+1\right)}}{1-k}$$ is never positive. It reaches a minimum value at $k=1+\sqrt 2$ and for such a value we have $-0.0470219$.
Just to be sure, I also computed the partial derivatives and set them equal to $0$. The problem is that this leads to $7$ solutions among which only two are explicit.
Running a few cases, it seems that the minimum effectively occurs at $x=k$ and $y=y_*$. So, to me the inequality does not hold.
I suggest you to do a contour plot of $f(x,y)$ for any $k$ of your choice and ask for the level $-0.01$. You will find here the case for $k=10$; the area on the left of or below the line corresponds to negative values of $f(x,y)$.
Edit
After comments, considering the inequality
$$\frac{x-k}{x+1}\times\frac{y-k}{y+1}+\frac{x-k}{x+1}+\frac{y-k}{y+1}\geq \color{red}{K}\frac{x y-k^2}{x y+1}$$ the problem is to find $K$.
So, as before, I considered the function
$$f(x,y)=\frac{x-k}{x+1}\times\frac{y-k}{y+1}+\frac{x-k}{x+1}+\frac{y-k}{y+1}- K\frac{x y-k^2}{x y+1}$$ then computed $f(x,x)$, assumed $x=k+\epsilon$, developed as a series. The constant term of the expansion is
$$-2 (k+1)^2 \left(k^2+1\right) (k(k+1)K-(1+k^2))$$ making it equal to $0$ gives
$$K=\frac{k^2+1}{k (k+1)}\implies f(x,x)=\frac{(k-x)^2 \,\,A}{k (k+1) (x+1)^2 \left(x^2+1\right)} $$
where
$$A=(2 k^2+3 k-1)x^2+2 \left(k^2-1\right) x+(k-1) (2 k+1)$$ does not show any real root. So, $f(x,x) \geq 0$
