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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 $|a-c|\leq 1, |b-d|\leq 1$?

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    @NicoBellic Any elements of $\mathbb{N}$.2012-05-19

1 Answers 1

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

$=-3(a^2+b^2)-2a(b+2)-4b-1<0$

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    Then interchange the roles of a,b,c,d in 3...!2012-05-19