Help with a proof: Given $a\ne 0$ and $b\ne 0$ and $a \lt \frac{1}{a} \lt b \lt \frac{1}{b}$. Prove $a\lt -1$ This question is from the book How to Prove It and I'm having trouble getting started with it.  The book provides the hint "first prove $a \lt 0$". However, I can't figure out how to get that far with the provided givens.
The question is:
Given $a\ne 0$ and $b\ne 0$ and $a \lt \frac{1}{a} \lt b \lt \frac{1}{b}$. Prove $a \lt -1$.
 A: HINT $\ \ \ $ Multiplying $\rm\ a < 1/a\ \:$ by $\rm\ a^2\ $ yields $\rm\:\ a^3 < a\ $. By symmetry $\rm\ b^3 < b\ $. So both $\rm\: a\:$ and $\rm \:b\:$ must lie in the intervals where the graph of $\ x^3 - x < 0\:,\ $ i.e. either in  $(-\infty,-1)$ or in $(0,1)\:$. But $\rm\ a \in (0,1)\ \Rightarrow\ 1/a > 1\ $ contra $\rm\ 1/a < b < 1\ $.

A: Suppose that $a$ were positive. Then, multiplying the string of ineqs. (doesn't change their direction) by $a$ gives $a^2 <1 < ba < a/b$. From here we see that $b>1$ (why?), but then, from the last ineq., $b^2 a < a$ which is impossible (why?). Therefore $a<0$. Now multiply again by $a$ and conclude.
A: $\ $ If $\rm\ a > 0\ $ then $\rm\ a < b\ $ times $\rm\ 1/a < 1/b\ \Rightarrow\ 1 < 1$   
So $\rm\ a < 0\ \:$ and $\rm\:\ a < 1/a\ \:\Rightarrow\:\ a^2 > 1 \ \:\Rightarrow\:\ a < -1$
A: Hint: Try to prove $a < 0$ by contradiction, i.e. assume $a > 0$ and try to derive a contradiction.
A: Hint: What does a<1/a say about a? What does b<1/b say about b?
A: Suppose $a\ge-1$.  Then adding $1$ to each term in the inequalities gives
$$0\le1+a\lt{1+a\over a}\lt1+b\lt{1+b\over b}$$
From $0\le1+a\lt{1+a\over a}$ and $0\lt1+b\lt{1+b\over b}$, it follows that $a$ and $b$ are positive, in which case
$${1+a\over a}\lt{1+b\over b}\implies b+ab\lt a+ab\implies b\lt a$$
But this contradicts the original inequality $a\lt b$.  Therefore we must have $a\lt-1$.
