Probability of One Event Less Than Probability of Second Event I am having a bit of trouble proving some cases when one probability is smaller than the other probability for all positive integers $a, b$, so some suggestions would be appreciated.
Here is the problem:

For all $a, b \in \mathbb{Z}^+$, if $P(A) = \dfrac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b}$ and $P(B) = \dfrac{a^2 + b^2}{a^2 + 2ab + b^2}$, prove that $P(A) < P(B)$.

So I attempted using cases.
Case 1: If $a = b \neq 0$, then we want to show that $P(A) < P(B)$
\begin{align*}
P(A) = \dfrac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b} = \dfrac{2a^2 - 2a}{4a^2 - 2a} < \dfrac{a^2 + b^2}{a^2 + 2ab + b^2} = \dfrac{2a^2}{4a^2} = \dfrac{1}{2} = P(B)
\end{align*}
Thus, this implies that $P(A) < P(B)$ for all $a, b \in \mathbb{Z}^+$
The next two cases are the cases I am having trouble with.
Case 2: If $a > b > 0$, then we want to show that $P(A) < P(B)$.
Case 3: If $b > a > 0$, then we want to show that $P(A) < P(B)$.
I am not sure how to approach cases 2 and 3. But case 3 should follow from case 2. So some assistance would be appreciated. Thanks
 A: I tried to make $P(A)$ and $P(B)$ comparable to avoid case decisions.
\begin{align}
 &P(A)=\frac{(a+b)^2-2ab-a-b}{(a+b)^2-a-b}=1-\frac{2ab}{(a+b)^2-a-b}=1-f(a,b)&&\\ \\
            & P(B)=\frac{(a+b)^2-2ab}{(a+b)^2}=1-\frac{2ab}{(a+b)^2}=1-g(a,b)&&
\end{align}
Then we start by comparing the denominators.
\begin{align}
&(a+b)^2-a-b<(a+b)^2 \\ \\  \Longrightarrow & f(a,b)>g(a,b)  \\ \\  \Longrightarrow & P(A)<P(B)
\end{align}
A: $$P(A) < P(B)\\
\iff \frac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b}<\frac{a^2 + b^2}{a^2 + 2ab + b^2}\\
\iff \frac{a^2+b^2-(a+b)}{(a+b)^2-(a+b)}-\frac{a^2+b^2}{(a+b)^2}<0\\
\iff \frac{(a^2+b^2-(a+b))(a+b)-(a^2+b^2)((a+b)-1)}{(a+b)^2((a+b)-1)}<0\\
\iff \frac{(a^2+b^2)(a+b)-(a+b)^2-(a^2+b^2)(a+b)+(a^2+b^2)}{(a+b)^2((a+b)-1)}<0\\
\iff \frac{-2ab}{(a+b)^2((a+b)-1)}<0\\
\iff \frac{ab}{(a+b)^2((a+b)-1)}>0.$$
For $a,b\in\mathbb Z^+,$
    $ab,\,(a+b)^2,\,((a+b)-1)$ are each positive (noting that $a+b\geq2$),
    so that last inequality is true,
    so $P(A)<P(B).$
A: Incomplete solution:
$$
P(A)=P(B)-\left[1-P(B)\right]\cdot\frac{a+b}{a^{2}+2ab+b^{2}-a-b}
$$
