I have a $V$ function:
$$V_n(x, y) = -\frac{1+2x}{7+5x+n+2nx} + \frac{-1+2x}{2-n+x(5+2n)} - \frac{(1-2y)^2}{(2-n+y(5+2n))^2} + \frac{4(2+y)^2}{(7+n+y(5+2n))^2}.$$
Here's the input form in case you want to copy it:
-((1 + 2 x)/(7 + n + 5 x + 2 n x)) + (-1 + 2 x)/(2 - n + (5 + 2 n) x) - (1 - 2 y)^2/(2 - n + (5 + 2 n) y)^2 + (4 (2 + y)^2)/(7 + n + (5 + 2 n) y)^2
This is strictly increasing in $x$ and strictly decreasing in $y$. I want to prove that $V_n(x, y) \geq 0$ implies $V_n(x+1, y+1) \geq 0$ for all $n$.
One of the ways I can think of is to define $f_n(x)$ to be $y$ that solves $V_n(x, y) = 0$. Then this is equivalent to proving that $f_n(x + 1) - f_n(x) \geq 1$. A sufficient condition is $f_n'(x) \geq 1$.
Plots of $f_{n=10}(x)$ (blue) and $f_{n=100}(x)$ (orange)
You can see that the function is well-behaved, but it's extremely complicate (the $V$ function not so much, but the $f$ function is). I've been stuck here a while now. Any ideas or thoughts would be much appreciated! If it doesn't involve the $f$ function, that's totally fine too.