AL be attitude, K,J,L be projection, MNP be medial triangle, D,E,F be tangent point of (I), D',E',F' be Nagel point. Notice ALD'~IDM so AD'/AL=IM/ID similarly for B so we just need AL.IM+BK.IN>pr or IM/MC+IN/NC>1. By Pytago, denote sqrt=', we left to '(4r^2+(b-c)^2/a^2) + '(4r^2+(a-c)^2/b^2)>1. Now if b-c/a + a-c/b > 1 or (a/b+b/a)>c(1/a+1/b) or (a^2+b^2)>c(a+b), if a+b>2c or c<min(a,b) then done. If C>=60, IM/MC+IN/NC=sinC/2(1/sinMIC+1/sinNIC)>2sinC/2>=1, done. Hence C<60 then c<max(a,b)=a wlog so b<c<a (isoceles is easy case). Now it equal to '((2br)^2+(bc-b^2)^2)+'((2ar)^2+(a^2-ac)^2)>ab. By Ptolemy, if IC.MN>MC.NC then done, it equivalent to sinC>=cos(A-B). If A-B>=90 then RHS just negative thus A-B<90 so sin(90-A+B)>sinC or 90>C+A-B=180-2B (sine function increasing on [0,90]) or B>45 but B<C<A then C in (45,60). Notice a+b<2c means c-b>a-c so we just need 4(4r^2+(a-c)^2)>b^2. It means 2IN>b/2, notice 4IN^2=2IA^2+2IC^2-AC^2 (median lemma), it equivalent to 2IA^2+2IC^2>5/4AC^2. Notice (sinC/2)^2 + (sinA/2)^2 > 5/8 > 5/8(cosB/2)^2, which is obviously true by LHS>(sin22.5)^2+(sin30)^2=1/2-'2/4+1/2>5/8<=>8 -2'2>5, thus we have the conclusion (very strictly)!