If the product of two +ve numbers is 256 then the least value of their sum is

$$\begin{gathered} Let two positive numbers be x,y \\ Given that xy=256\Rightarrow y=\frac{256}{x} \\ Now x+y=x+\frac{256}{x} \\ let f\left(x\right)=x+\frac{256}{x} \\ we know that A.M\ge G.M \\ \Rightarrow \frac{x+\frac{256}{x}}{2}\ge \sqrt{x\left(\frac{256}{x}\right)} \\ \Rightarrow x+\frac{256}{x}\ge 2\left(16\right) \\ \Rightarrow f\left(x\right)\ge 32 \\ \therefore minimum value of f\left(x\right)=32 \end{gathered}$$