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.



Inequality involving product-of-minus vs minus-of-product for positive integers
We need to prove that
$$(a\_1a\_2-b\_1b\_2)(a\_1-1)(a\_2-1)\geq(a\_1-b\_1)(a\_2-b\_2)(a\_1a\_2-1),$$ which is a linear inequality of $b\_1$ and of $b\_2$ and since
$$1\leq b\_1\leq a\_1-1$$ and $$1\leq b\_2\leq a\_2-1,$$ it's enough to prove our inequality for $b\_1\in\{1,a\_1-1\}$ and $b\_2\in\{1,a\_2-1\},$ 


where $a\_1\geq2$ and $a\_2\geq2$.


1. $b\_1=b\_2=1.$


We obtain an identity;


2. $b\_1=1$, $b\_2=a\_2-1$.


We need to prove that
$$(a\_1a\_2-a\_2+1)(a\_2-1)\geq a\_1a\_2-1$$ or
$$a\_2(a\_2-2)(a\_1-1)\geq0,$$ which is obvious;


3. $b\_1=a\_1-1$ and $b\_2=1$.


This case a similar to the previous case;


4. $b\_1=a\_1-1$ and $b\_2=a\_2-1$.


We need to prove that
$$(a\_1+a\_2-1)(a\_1-1)(a\_2-1)\geq a\_2a\_2-1$$ or
$$(a\_1+a\_2-2)(a\_1a\_2-a\_1-a\_2)\geq0$$ or
$$(a\_1+a\_2-2)((a\_1-1)(a\_2-1)-1)\geq0.$$
Done!


