Proving if $x _{1}^{2} - y _{1}^{2} = -1, x _{2}^{2} - y _{2}^{2} = -1, y_1y_2 > 0$ then $(x_2 - x_1)^2 - (y_2 - y_1)^2 > 0$?

Question

I just expand $$(x_2 - x_1)^2 - (y_2 - y_1)^2 = (x _{2}^{2} - y _{2}^{2}) + (x _{1}^{2} - y _{1}^{2}) - 2x_1x_2 + 2y_1y_2 \\ = -2 -2x_1x_2 + 2y_1y_2,$$ and I'm stuck here.

Intuitively, the graph of points $(x,y)$ satisfying $x^2 - y^2 = -1$ supports the conclusion. Here is the graph

Answer 1

I'm assuming that $x_2 > x_1 \geq 0$. Deal with the negative $x_i$ cases in a similar manner.

We are given that $ y_i = \sqrt{ 1 + x_i^2 }$.
Then, we WTS

$$ | x_2 - x_1 | > \left| \sqrt{ 1 + x_2^2 } - \sqrt{ 1 + x_1 ^2 } \right|. $$

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Approach 1:
WLOG $x_2 > x_1$.
$0 < \sqrt{ 1 + x_2 ^2 } - \sqrt{ 1 + x_1^2 } = \frac{ x_2 ^2 - x_1 ^2 } { \sqrt{ 1 + x_2^2 } + \sqrt{ 1 + x_1 ^ 2 } } < \frac{ x_2^2 - x_1 ^2 } { x_2 + x_1 } = x_2 - x_1 $.

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Approach 2:

WLOG, $x_2 > x_1$.
Proof by contradiction. Suppose not, then
$x_2 - x_1 \leq \sqrt{1 +x_2^2 } - \sqrt{ 1 + x_1^2}$.
$x_2 - x_1 (\sqrt{1 +x_2^2 } + \sqrt{ 1 + x_1^2}) \leq x_2 ^2 - x_1 ^2 = (x_2 - x_1) ( x_2 + x_1) $
$\Rightarrow \sqrt{1 +x_2^2 } + \sqrt{ 1 + x_1^2} \leq x_2 + x_1$

Adding these 2 inequalities, we get
$ \sqrt{ 1 + x_1^2 } \leq x_1 $, which is a contradiction.

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Approach 3:

Hint: Show (E.g. by differention) that the gradient of $ \sqrt{1 + x^2}$ has absolute value strictly less than 1.

Answer 2

Because $$(x_1-x_2)^2-(y_1-y_2)^2=2(y_1y_2-x_1x_2)=2(\sqrt{(x_1^2+1)(x_2^2+1)}-x_1x_2)>$$ $$>2(\sqrt{x_1^2x_2^2}-x_1x_2)=2(|x_1x_2|-x_1x_2)\geq0.$$