Let $f(a,b,c,d)=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}-\sqrt{(a+c)^2+(b+d)^2}$. (it is the defect in the triangle inequality)

Then, we discovered by heuristic arguments and then verified by computer that $$\sum f(a,b,c,d)^n = 2-\pi/2$$ where the sum runs over all $a,b,c,d\in\mathbb Z$ such that $a\geq 1,b,c\geq 0, ad-bc=1$ and $n=2$.

It seems that when $n=1$, we obtain $2$ in the right hand side. We have failed to guess the result for $n>2$.

So the question is: can you prove the result for $n=2$? (*We can, but in a rather unnatural way. We will write this later and now we want to tease the community, probably somebody can find a beautiful proof.*)

Added: we have two answers for the above question, so the rest is:

**Question:** Can you guess the result for $n>2$? I tried http://mrob.com/pub/ries/ but nothing interesting was revealed.

PS. In case it can help someone, below are these sums of powers ($n=1,2,3,4,5$), calculated by computer:

$1.9955289122768913 = 2$

$0.4292036731309361 = 2 - \pi/2$

$0.21349025954227965 = $ ?

$0.11983665032283052 = $ ?

$0.06933955916793563 = $ ?

Added: this and something more can be found in https://arxiv.org/abs/1701.07584 and https://arxiv.org/abs/1711.02089