The least natural number $a$ for which  $x+a{x}^{-2}>2 \mathrm{\forall }x\in (0,\mathrm{\infty })$  is

Let  $$\begin{gathered} x+a{x}^{-2}>2 \\ a>\frac{(2-x)}{x^{-2}}\Rightarrow a>x^2(2-x) \\ a=2x^2-x^3 \\ a\text{'}=4x-3x^2 \\ a\text{'}=0 \\ \Rightarrow x=0, \frac{4}{3} \\ a"=4-6x \\ minimum at x=\frac{{\displaystyle 4}}{{\displaystyle 3}} \\ a\left(\frac{{\displaystyle 4}}{{\displaystyle 3}}\right)=\frac{32}{27}\Rightarrow \text{ a has to be greater than }\frac{32}{27} \end{gathered}$$, 

Use the concept of Maxima & minima