Question

The least value of $\alpha \in R$ for which $4 \alpha x^{2}+\frac{1}{x} \geq 1$, for all $x>0$, is -

• A. $\frac{1}{64}$
• B. $\frac{1}{32}$
• C. $\frac{1}{27}$
• D. $\frac{1}{25}$

Answer 1

Answer: (C)

$\frac{4 \alpha x^{2}+\frac{1}{2 x}+\frac{1}{2 x}}{3} \geq\left(\frac{4 \alpha x^{2}}{4 x^{2}}\right)^{1 / 3} $
( where $\alpha>0$ obviously)
$4 \alpha x^{2}+\frac{1}{x} \geq 3 \alpha^{1 / 3}$
$\therefore 3 \alpha^{1 / 3} \geq 1$
$\alpha \geq \frac{1}{27}$