Q.457. Trigonometric inequality.

Question 457.
Prove that (sin^3 A)/(sin B) + (cos^3 A)/(cos B) ≥ sec ( A - B ) for all 0 < a,b < π/2.

Answer 457.
(sin^3 A)/(sin B) + (cos^3 A)/(cos B) ≥ sec ( A - B )
<=> (sin^3 A cosB + cos^A sinB) / sinB cosB ≥ 1/cos(A - B)
<=> (sin^3 A * cosB + cos^3 A * sinB) * cos(A - B) ≥ sinB cosB
<=> (sin^2 A * 2sinA cosB + cos^2 A * 2cosA sinB) * cos(A - B) ≥ 2sinB cosB
<=> [sin^2 A {sin(A+B) + sin(A-B)} + cos^2 A {sin(A+B) - sin(A-B)}] * cos(A - B) ≥ sin2B
<=> [(sin^2 A + cos^2 A) sin(A + B) - (cos^2 A - sin^2 A) sin(A - B)] * cos(A - B) ≥ sin2B
<=> sin(A + B) cos(A - B) - cos2A sin(A - B) cos(A - B) ≥ sin2B
<=> 2sin(A + B) cos(A - B) - cos2A * 2sin(A – B) cos(A - B) ≥ 2sin2B
<=> sin2A + sin2B - cos2A * 2sin(A – B) cos(A - B) ≥ 2sin2B
<=> sin2A - sin2B - cos2A * 2sin(A – B) cos(A - B) ≥ 0
<=> 2cos(A + B) sin(A - B) - cos2A * 2sin(A - B) cos(A - B) ≥ 0
<=> sin(A - B) [cos(A + B) - cos2A cos(A - B)] ≥ 0
<=> sin(A - B) [2cos(A + B) - 2cos2A cos(A - B)] ≥ 0
<=> sin(A - B) [2cos(A + B) - cos(3A - B) - cos(A + B)] ≥ 0
<=> sin(A - B) [cos(A + B) - cos(3A - B)] ≥ 0
<=> 2sin2A sin^2 (A - B) ≥ 0
which is true
=> the given inequality is true.

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