I'm encountering this inequality for dimensionality reduction problem. The simplified form looks as follows:

Consider positive integers $a_1$, $a_2$, $b_1$ and $b_2$ where $a_1>b_1$ and $a_2>b_2$. Prove that

$$ \frac{a_1a_2-b_1b_2}{a_1a_2-1}\geq\frac{(a_1-b_1)(a_2-b_2)}{(a_1-1)(a_2-1)} $$

The inequality seems very trivial and easy but I am struggling to prove it. While I could prove for the special cases where (1) $a_1=a_2=a$, which reduces to

$$ (a-1)[(b_1+b_2-2)a-(2b_1b_2-b_1-b_2)]\geq0 $$

$$ \iff a\geq max(b_1,b_2)\geq\frac{b_1(b_2-1)+b_2(b_1-1)}{(b_2-1)+(b_1-1)}, $$

and (2) $b_1=b_2=b$, which reduces to

$$ (a_1a_2+b)(a_1+a_2)\geq a_1a_2(2b+2), $$

I cannot verify the general case where $a_1\neq a_2$ and $b_1 \neq b_2$. If someone could help to provide guidance, reference to similar inequalities in the literature, or any idea to the solution, I would be very thankful.