Q.36. Inequality of Trigonometry

Question 36.
Let a, b and c be the sides of an acute triangle with a circumradius of R.
Show that R^6 ≥ (4R² - a²)(4R² - b²)(4R² - c²).

Answer 36.
R^6 ≥ (4R² - a²)(4R² - b²)(4R² - c²)
<=> R^6≥(4R² - 4R^2sin^2 A)(4R² - 4R^2sin^2 B)(4R² - 4r^2 sin^2 C)
<=> 1 ≥ 64(1 - sin^2 A)(1 - sin^2 B)(1 - sin^2 C)
<=> 1/64 ≥ cos^2 A * cos^2 B * cos^2 C
<=> 1/8 ≥ lcosA cosB cos Cl
<=> 1 - 8 lcosA cosB cosCl ≥ 0
<=> 1 - 8cosAcosBcosC ≥ 0 [for acute triangle]
<=> 1 - 4(2cosAcosB)cosC ≥ 0
<=> 1 - 4[cos(A+B) + cos(A-B)]cosC ≥ 0
<=> 1 - 4[cos(π-C) + cos(A-B)]cosC ≥ 0
<=> 1 - 4[-cosC + cos(A-B)]cosC ≥ 0
<=> 4cos^2 C - 4cosCcos(A-B) + 1 ≥ 0
<=> [2cosC - cos(A-B)]^2 + 1 - cos^2 (A-B) ≥ 0
<=> [2cosC - cos(A-B)]^2 + sin^2 (A-B) ≥ 0
which is true
=> original inequality is true for acute triangle.

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