I have two progressions: $$\begin{align}P_1&: xy, xy + (x+2)(y+2), xy + (x+2)(y+2) + (x+4)(y+4),\dots, \sum_1^i (x+2(j-1))(y+2(j-1))\end{align}$$ $$\begin{align}P_2&: (x+2)(y+2), (x+4)(y+4),\dots, (x + 2i)(y+2i)\end{align}$$
$x, y$ are natural numbers.
I just want to know if the ratio $P_1(i):P_2(i)$ is an increasing function. It appears to be. But how to prove it?
I would like to add that $x > 3$, $y > 3$ are odd numbers.
Term $k$ of $P_1$ is $$[x+2(k-1)][y+2(k-1)]=xy+2(k-1)(x+y)+4(k-1)^2$$ Summing by terms from $k=1..n$ we get $$P_1(n)=nxy+n(n-1)(x+y)+\frac23(2n-1)n(n-1)$$ Meanwhile $$P_2(n)=(x+2n)(y+2n)=xy+2n(x+y)+4n^2$$
Therefore as $n \to \infty$ then $$\frac{P_1(n)}{P_2(n)} \to \frac14(x+y)+\frac13 n$$ which continues increasing indefinitely for all values of $x, y$.