Question

Show that 2sinx + tanx ≥ 3x, where 0 ≤ x < π/2  

Answer 1

Let f(x) = 2sinx + tanx – 3x

f'(x) = 2cosx + sec²x – 3

⇒ f'(x) = (cosx + cosx + sec²x) – 3

⇒ f'(x) ≥ 3 – 3 = 0 ( A.M ≥ G.M)

⇒ f'(x) ≥ 0 f(x) is strictly increasing function

Thus, x ≥ 0 ⇒ f(x) ≥ f(0)

⇒ 2sinx + tanx – 3x ≥ 0

⇒ 2sinx + tanx ≥ 3x