# 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 $$k=1.$$

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

• 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! – mathlover Apr 19 at 10:01
• @mathlover. I suppose. Now, the problem is to find $K$. Cheers :-) – Claude Leibovici Apr 19 at 10:05
• Great. Thanks for all the suggestions. – mathlover Apr 19 at 10:06
• @mathlover. Have a look at the edit. – Claude Leibovici Apr 19 at 11:22
• excellent modifications after comments. Thanks. – mathlover Apr 19 at 11:33