# Inequality involving product-of-minus vs minus-of-product for positive integers

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.

• It is linear in $b_1$, so it suffices to consider $b_1=1$ and $b_1=a_1-1$. The same for $b_2$. – Fedor Petrov Feb 18 '19 at 14:10
• Thank you for the explanation! @FedorPetrov – Piccadilly Dough Feb 19 '19 at 17:14

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;

1. $$b_1=1$$, $$b_2=a_2-1$$.

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;

1. $$b_1=a_1-1$$ and $$b_2=1$$.

This case a similar to the previous case;

1. $$b_1=a_1-1$$ 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!

• Thank you so much for your detailed explanation of proof. I really appreciate it! – Piccadilly Dough Feb 19 '19 at 17:15
• You are welcome! – Michael Rozenberg Feb 19 '19 at 18:53