JEE Advanced 2016 · Paper 1, Mathematics question 41

AM-GM. Split $\frac1x$ into two equal halves:
$$4\alpha x^2+\frac1{2x}+\frac1{2x}\ge3\sqrt[3]{4\alpha x^2\cdot\frac1{4x^2}}=3\alpha^{1/3}.$$
Equality is reached at some $x>0$, so this is the minimum.

Condition. $3\alpha^{1/3}\ge1$, so $\alpha\ge\dfrac1{27}$.

(C)