Relationship between a and b based on conditions Suppose that you have two real numbers $a$ and $b$. These numbers have the property that there exists a real number $c>0$ such that $a+c=b$, and for all $d$, $-1<d<0$, $da>b$. Is $a^2>b^2$, $a^2<b^2$, can both be possible, or are none of the choices certain?
From the second condition with $d$ I inferred that $a$ and $b$ must be negative, since $d$ can be brought arbitrarily close to 0, yet $da$ will still be greater than $b$. Then, from the first condition with $c$, I concluded that $a$ must be less than $b$, as there is a positive difference between $a$ and $b$. Therefore, $a^2 > b^2$. Is my conclusion correct?
 A: Yes. Indeed, the answer is: $a^2>b^2$.
We have,
$$a^2-b^2=-c(a+b)$$
Then, we need only the sign of $a+b$.
Since,
$$a(d-1)>0\implies a<0$$
This implies that,
$$0>ad+a=a(d+1)>b+a$$
Therefore,
$$a^2-b^2=-c(a+b)>0.$$
A: Minor nitpick.  If you don't know that $a < b$ then knowing that for all $d: -1 < d < 0$ yields $da > b$ can't allow you to conclude that $a$ and $b$ are both negative.  So $d$ may be arbitrarily close to $0$ then $da$ is either equal to $0$ if $a$ is, or arbitrarily close to $0$ (but we don't know if $a$ is pos., neg. or $0$ so we don't know if $da < 0$ or $da=0$ or $da >0$) either way, if $b < ad$ for all such $d$ no matter how small $|d|$ is we must have $b\le 0$.
But we could have $b = 0$ (If $a \le 0$).  But as we don't know $a < b$ we can't conclude anything about $a$.  We can conclude that if $a < 0$ then $b\le 0$.  If $a=0$ then $b< 0$. And if $a >0$ then $b< 0$.
But that is all we can conclude from statement 2 alone.
....
but of course we have statement 1.  Statement 1 is nothing more or less than simply stating:  $a < b$.   With that in mind, and your reasoning of statement 2 we get $a < b \le 0$.  But again.  $b =0$ is a possiblity that satisfies both statement $1$ and $2$.
BTW... I reasoned with statement 2 (and having concluded $a< b$ by statement 1) that $ad > b\implies a < \frac bd$.  But as $d < 0$ then $b$ and $\frac bd$ are "opposite signs" meaning at least one of them is less or equal to $0$ so $a < 0$. .... but I guess that wasn't nesc. as the "arbitrary close to $0$" would have told me that $b \le 0$ had I thought to do that argument at the time.
......
Anywhoo.... $a < b \le 0 \implies b^2 < a^2$
A: It is not correct that the second condition alone implies that both $a$ and $b$ must be negative.
The second condition implies that $b\leq0$, since
$$\inf~\{da:-1<d<0\} \leq 0$$
for all $a \in \mathbb{R}$.
On the other hand, the second condition also implies that $(-d)a<-b$ for all $0<-d<1$ and thus $a<-b$. This means that from the second condition only, $a$ may be positive.
The first condition is just a way of writing $a<b$, so $a<b\leq 0$ and thus $a$ is indeed negative and consequently $a^2>b^2$.
