The least positive integral value of ‘a’ such that 2x+$\frac{a}{x^2}\ge 6, \forall x\in R is$

$\text{Let f(x) =2x+}\frac{a}{x^2}-6$

$We have to find \text{'}a\text{'} when 2x+\frac{a}{x^2}-6\ge 0, \forall x\in R$

$\Rightarrow we have to ensure that minimum value\left(a\right)of f\left(x\right) ⩾0$

$\therefore f^{\text{'}}\left(x\right) =2-\frac{2a}{x^3} and f^{"}\left(x\right)=\frac{6a}{x^4}$

$Also f^{"}\left(a^{\frac{1}{3}}\right)>0$

$f^{\text{'}}\left(x\right) =0 \Rightarrow x=a^{\frac{1}{3}}$

$hence f\left(x\right) is maximum at x=a^{\frac{1}{3}}$

$\therefore f\left(x\right) \ge 0 if 2a^{\frac{1}{3}}+ \frac{a}{a^{{\displaystyle \frac{2}{3}}}}-6 \ge 0 , \forall x\in R$

$$\begin{gathered} \therefore a^{\frac{1}{3}} \ge \frac{6}{3} \\ a\ge 8 \end{gathered}$$