If $a^2 - b^2 < c^2 - d^2$ with these conditions (see body), necessarily imply that $a - b < c - d$?

Question

The conditions are the following:

1. $a > b$ and $c > d$
2. $a, b, c ~\text {and}~d$ are positive real numbers
3. $a > c, b > d$

I have tried to use triangle inequalities, but I haven't been able to find a proof yet.

Answer 1

Yes, $a^2 - b^2 < c^2 - d^2$ and $a+b > c+d > 0$ (from conditions 2 and 3) imply that $$ a-b = \frac{a^2-b^2}{a+b} < \frac{c^2-d^2}{c+d} = c-d \, . $$

Answer 2

Trick is to realize that $a^2 -b^2 = (a+b)(a-b)$ and $c^2 - d^2 =(c+d)(c-d)$

So $(a+b)(a-b) < (c-d)(c+d)$ and as$a,b>0$ then $a+b>0$ Then $a-b < (c-d)\frac {c+d}{a+b}$.

It is necessary and sufficient to show $\frac {c+d}{a+b} \le 1$.