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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 $k=1.$

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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, $f(x,x) \geq 0$

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  • $\begingroup$ Thanks. It appears that if the right hand side of the inequality is $\frac{1}{K} ((xy−k^2)/(xy+1)),$ the inequality might be true! $\endgroup$ – mathlover Apr 19 at 10:01
  • $\begingroup$ @mathlover. I suppose. Now, the problem is to find $K$. Cheers :-) $\endgroup$ – Claude Leibovici Apr 19 at 10:05
  • $\begingroup$ Great. Thanks for all the suggestions. $\endgroup$ – mathlover Apr 19 at 10:06
  • $\begingroup$ @mathlover. Have a look at the edit. $\endgroup$ – Claude Leibovici Apr 19 at 11:22
  • $\begingroup$ excellent modifications after comments. Thanks. $\endgroup$ – mathlover Apr 19 at 11:33

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