MathLabs

Problem 3

Let ABCABC be a triangle. The bisector of angle AA meets segment BCBC at XX and the circumcircle at YY. Let rA=AX/AYr_A=AX/AY, and define rB,rCr_B,r_C similarly. Prove that rA/sin⁡2A+rB/sin⁡2B+rC/sin⁡2C≥3r_A/\sin^2A+r_B/\sin^2B+r_C/\sin^2C\ge3, with equality if and only if the triangle is equilateral.
Step 3 of 5: Separate the product-one factors
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sA=sin⁡Bsin⁡Csin⁡2A,sAsBsC=1s_A=\frac{\sin B\sin C}{\sin^2A},\qquad s_As_Bs_C=1
Detailed analysis

Thus rA/sin⁡2A=sA/sin⁡2(B+A/2)r_A/\sin^2A=s_A/\sin^2(B+A/2), cyclically. The definitions give sAsBsC=1s_As_Bs_C=1, so AM-GM gives sA+sB+sC≥3s_A+s_B+s_C\ge3.