For $s \gt 2R + r$, please see Circumradius, Inradius section in en.wikipedia.org/wiki/Acute_and_obtuse_triangles

In the link you given one can find only the next reference and this other reference reads: "1088^⋆.*Proposed by Basil C. Rennie, James Cook University of North Queensland, Australia.* If $R, r, s$ are the circumradius, inradius, and semiperimeter, respectively, of a triangle with largest angle $A$, prove or disprove that $s\, ?\, 2R + r$ according as $A\,?\, 90^\circ$.

@user ok, pls see Section Inradius, exradii, and circumradius in en.wikipedia.org/wiki/List_of_triangle_inequalities and link#35 at the bottom

I see. You probably mean the Lemma 2 from the link#35.

Lemma 2 proves exactly this identity. :)

I think you should mention this identity in the answer because it is not so well-known.

@MathLover You're back..thank you for the answer

@petaarantes you are welcome!

3 > OP+OQ+OR = 2(OP+OQ+OR)/2 > (PQ+QR+RP)/2 = (AB+BC+CA)/4 = 2 ?

@Anonymous not sure I follow you. Can you please elaborate?

By the triangle inequality and the mid-pt theorem, OP+OQ>PQ and PQ = AB/2. Then, 2 < OP+OQ+OR < 3 ??

$OP + OQ \gt PQ$ yes and $PQ = AB/2$ yes but how does that lead to $OP + OQ + OR \gt 2$? All we know is $OP + OQ + OR \gt OR + AB/2$

I'm not sure whether I'm right or not. We can use the triangle inequality (3 times) and mid-pt theorem (3 times). OP+OQ+OR = 2(OP+OQ+OR)/2 = [(OP+OQ) + (OQ+OR) + (OR+OP)]/2 > (PQ + QR + RP)/2 = (AB/2 + CA/2 + BC/2)/2 = (AB+BC+CA)/4 = 8/4 = 2

you are absolutely right and that is a very good catch

@petaarantes please see comments from Anonymous. There is a stronger lower bound on the value of $OP + OQ + OR$ than what I have in my answer. So it seems there is no integer value possible for $OP + OQ + OR$. Can you please unaccept my answer? I would like to delete it.

@Anonymous you may want to add an answer showing no integer value is possible.