Question

If ax²+$\frac{b}{x}$≥ c for all positive real values of x. Then

• A. 27 ab²≥ 4 c³
• B. 27ac²≥ 4b³
• C. 27 bc²≥ 4a³
• D. All are correct

Answer 1

Answer: (A)

27 ab²≥ 4 c³

Answer 2

f(x) = ax² + $\frac{b}{x}$ – c ≥ 0

So that minimum value of $\left( ax^{2} + \frac{b}{x} - c \right)$ should be ≥ 0

f(x) = ax² + $\frac{b}{x}$ – c

f '(x) = 2ax – b/x² f '(x) = 0 ⇒ x = $\left( \frac{b}{2a} \right)^{1/3}$and

f "(x) = 2a + $\frac{2b}{x^{3}}$ > 0 if x = $\left( \frac{b}{2a} \right)^{1/3}$

∴ f(x) has minima at x = $\left( \frac{b}{2a} \right)^{1/3}$

f(x) ≥ 0,$f\left( \left( \frac{b}{2a} \right)^{1/3} \right)$ ≥ 0

a$\left( \frac{b}{2a} \right)^{\frac{2}{3}}$ + $\frac{b}{\left( \frac{b}{2a} \right)^{\frac{1}{3}}}$– c ≥ 0

a $\left( \frac{b}{2a} \right)$ + b ≥ c $\left( \frac{b}{2a} \right)^{1/3}$

$\left( \frac{3b}{2} \right)^{3}$ ≥ $\frac{b}{2a}$ c³ ⇒  27ab² ≥ 4c³_.