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.

  • 2
    $\begingroup$ It is linear in $b_1$, so it suffices to consider $b_1=1$ and $b_1=a_1-1$. The same for $b_2$. $\endgroup$ Commented Feb 18, 2019 at 14:10
  • $\begingroup$ Thank you for the explanation! @FedorPetrov $\endgroup$ Commented Feb 19, 2019 at 17:14

1 Answer 1


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!

  • $\begingroup$ Thank you so much for your detailed explanation of proof. I really appreciate it! $\endgroup$ Commented Feb 19, 2019 at 17:15
  • $\begingroup$ You are welcome! $\endgroup$ Commented Feb 19, 2019 at 18:53

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