Take the 2-minute tour ×

How to prove this inequality:

$$a^2 d^2+b^2 c^2-1-4ac-4bd-2abcd \leq 0,$$

where: $a, b, c, d \in \{0, 1, 2, \ldots\}$ and $|a-c|\leq 1, |b-d|\leq 1$?

share|improve this question
What do you mean by $a,b,c,d=0,1,2,\ldots$? If these are known you can just insert them and see what happens. Otherwise you might possibly mean $\in \mathbb{N}$? –  user20266 May 19 '12 at 14:22
Of course, I mean $a,b,c,d \in \{\mathbb(N) \cup 0\}$ –  Sh.N. May 19 '12 at 14:26
So is $d=3$? or $a,b,c,d$ are any elements of $\mathbb{N}$? –  Nico Bellic May 19 '12 at 14:36
@NicoBellic Any elements of $\mathbb{N}$. –  Sh.N. May 19 '12 at 14:46

1 Answer 1

up vote 1 down vote accepted

If I understood your question correctly, all the numbers are non-negative integers and either $\,a=c\,\,or\,\,a=c\pm 1\,$ , and the same for $\,b,d$ .

Now, put $\,\omega:=\,a^2d^2+b^2c^2-1-4ac-4bd-2abcd=(ad-bc)^2-(4ac+4bd+1)$ , so:

1) Suppose $\,a=c\,,\,b=d\Longrightarrow \omega=-4(a^2+b^2)-1<0$ ;

2) Suppose $\,a=c\pm 1\,,\,b=d\Longrightarrow \omega=d^2-4d^2-4c^2+4c-1=-3d^2-4\left(c-\frac{1}{2}\right)^2<0$

3) Suppose $c=a + 1\,,\,\,d=b+ 1\,\Longrightarrow \omega=(a-b)^2-4a(a+1)-4b(b+1)-1=$


share|improve this answer
And if $c = a-1, d= b-1$? –  Sh.N. May 19 '12 at 14:35
Then interchange the roles of a,b,c,d in 3...! –  DonAntonio May 19 '12 at 14:44

Your Answer


