an elementary inequality for arbitrary real numbers

Question

Does anyone happen to recognize the elementary inequality shown in the image? It appeared in a journal paper and can have a general form, but unfortunately not one that I recognize as it stands. (This is my first post and hopefully I uploaded the image correctly; apologies if I failed.)

Answer 1

$$|d(0, (a,b)) - d(0, (c,d))| \le d((a,b),(c,d)| \le d((a,b),(c,b)) + d((c,b),(c,d))$$

Answer 2

The inequality holds if and only if the inequality between the squares hold. This last inequality is$$a^2+b^2+c^2+d^2-2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\leqslant a^2+b^2+c^2+d^2-2ac-2bd,$$which is equivalent to$$ac+bd\leqslant\sqrt{a^2+b^2}\sqrt{c^2+d^2},$$a trivial consequence of the Cauchy-Schwarz inequality.

Answer 3

We need to prove that $$a^2+b^2-2\sqrt{(a^2+b^2)(c^2+d^2)}+b^2+d^2\leq(a-c)^2+2|(a-c)(b-d)|+(b-d)^2$$ or

$$\sqrt{(a^2+b^2)(c^2+d^2)}\geq ac+bd-|(a-c)(b-d)|,$$ which follows from C-S: $$\sqrt{(a^2+b^2)(c^2+d^2)}\geq ac+bd\geq ac+bd-|(a-c)(b-d)|$$ and we are done!