1) Let R be the radius of circumcircle of triangle ABC
We know that in a triangle ABC:
1) cosA+cosB+cosC=1+r/R
2) sinA+sinB+sinC=s/R
3) R≥2r
Square and Add equations (1) and (2) and use
sin²A+cos²A=1
=> 3+2(cosAcosB+sinAsinB+cosBcosC+sinBsinC+cosCcosA+sinCsinA) = 1+2r/R+(r/R)²+(s/R)²
Now substitute ineq(3)
=> 3+2(cos(A-B)+cos(B-C)+cos(C-A)) ≤ 1+1+1/4+(s²/4r²)
=> 8r²(cos(A-B)+cos(B-C)+cos(C-A)) ≤ s²-3r²
2) Let R be radius of circumcircle, s be the semiperimeter and r be inradius of triangle of ABC.
=> 1/sinAsinB+1/sinBsinC+1/sinCsinA
=4R²(1/ab+1/bc+1/ca)
=4R²(a+b+c)/abc
But Area =abc/4R=rs and 2s=a+b+c
So, 1/sinAsinB+1/sinBsinC+1/sinCsinA = 2sR/Area
= 2sR/rs = 2R/r ≥ 4 (by ineq(3)) in prob (1)
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