I have this statement:
Let $a, b, c, d \in \mathbb{R} - $ {$0$}, with $\quad acd> 0$.
If $– 1 < \frac{a}{b} < \frac{b}{c} < \frac{c}{d} < \frac{d}{a} < 1,$
Which of the following alternative are false?
A) $\quad a < cd$
B) $\quad c < ad$
C) $\quad d < ac$
D) $\quad ab < cd$
E) $\quad ad < bc$
I'm really stuck, I've tried several things and nothing has worked for me. Some hint or guide would be useful
On
Check where $0$ can fit ($5$ options): $$– 1 \overbrace{<}^{<0} \frac{a}{b} \overbrace{<}^{<0} \frac{b}{c} \overbrace{<}^{<0} \frac{c}{d} \overbrace{<}^{<0} \frac{d}{a} \overbrace{<}^{<0} 1,$$ For each case, by checking signs of the numbers $a,b,c,d$, one can find only the central sign works. For example: $$– 1 <0< \frac{a}{b} < \frac{b}{c} < \frac{c}{d} < \frac{d}{a} < 1 \Rightarrow \\ a<b<c<d<a \ ?$$ Similarly, you can reject the rest three cases by contradiction.
So we have: $$– 1 < \frac{a}{b} < \frac{b}{c} < 0<\frac{c}{d} < \frac{d}{a} < 1 \Rightarrow \\ 1) \ (a,b,c,d)=(+,-,+,+); \ \ 2) \ (a,b,c,d)=(-,+,-,-) \Rightarrow \\ 1) \ E) \ ad<bc \iff (+)(+)<(-)(+) \ ? \ 2) \ E) \ ad<bc \iff (-)(-)<(+)(-) \ ?$$ I leave other options to check to you.
Assuming by "different of zero" you mean $abcd\neq 0$.
We have $a/c=(a/b)(b/c)$ so $-1<a/c<1$, and we are also given $-1<c/d<1$, $-1<d/a<1$. Multiplying gives the contradiction $1=(acd)/(acd)<1$ hence there are no $a,b,c,d$ that satisfy the given inequality. Hence all options are vacuously true.