How to prove this \(\displaystyle{\frac{{{\sin{{\left({A}-{B}\right)}}}{\sin{{\left({A}-{C}\right)}}}}}{{{\sin{{2}}}{A}}}}+{\frac{{{\sin{{\left({B}-{C}\right)}}}{\sin{{\left({B}-{A}\right)}}}}}{{{\sin{{\left({2}{B}\right)}}}}}}+{\frac{{{\sin{{\left({C}-{A}\right)}}}{\sin{{\left({C}-{B}\right)}}}}}{{{\sin{{2}}}{C}}}}\geq{0}\)

Breanna Mcclure

Breanna Mcclure

Answered question

2022-04-08

How to prove this
sin(AB)sin(AC)sin2A+sin(BC)sin(BA)sin(2B)+sin(CA)sin(CB)sin2C0

Answer & Explanation

chambasos6

chambasos6

Beginner2022-04-09Added 12 answers

sin2Bsin2Csin(AB)sin(AC)=sin(2(πAC))sin(2πAB))sin(AB)sin(AC)
=4[sin(A+C)cos(A+C)][sin(AC)cos(AC)]sin(A+B)sin(AB)
=cos(B)cos(C)[sin2(A)sin2(B)][sin2(A)sin2(C)]
Using the extended Schur's Inequality
cos(B)cos(C)[sin2(A)sin2(B)][sin2(A)sin2(C)]0

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