# Prove that sinx + 2x ≥ 3x (x+1)/π ∀x ∈[0 , π/2]. (Justify the inequality,if any used).

Q: Prove that $\displaystyle sinx + 2x \ge \frac{3x(x+1)}{\pi}$ ∀ x ∈ [0 , π/2]. (Justify the inequality,if any used).

Sol: Let f (x) = 3x2 + (3 – 2π ) x – π sin x

f (0) = 0, f (π/2) = – ve

f ‘(x) = 6x + 3 – 2π – π cos x

f ” (x) = 6 + π sin x > 0

⇒ f ‘ (x) is increasing function in [0 , π/2]

⇒ there is no local maxima of f(x) in [0 , π/2]

⇒ graph of f(x) always lies below the x-axis in [0 , π/2]

f(x) ≤ 0 in x ∈ [0 , π/2]

3x2 + 3x ≤ 2πx + π sinx

$\displaystyle sinx + 2x \ge \frac{3x(x+1)}{\pi}$