10-seconds challenge-1

Using $AM\geq GM$

$\frac{x+\frac{1}{x}}{2}$ $\geq$ $\sqrt{x.\frac{1}{x}}$

$\implies$ $\frac{x+\frac{1}{x}}{2}$ $\geq$ $1$

$\implies$ $x+\frac{1}{x}$ $\geq$ $2$

equality holds when $x=\frac{1}{x}$ $\implies$ $x=\pm 1$

But $x>0$ and also $x \neq 1$

$\implies$ $min$ ( $x+\frac{1}{x}$ ) $\neq$ $2$

So the range of given funtion is $(2,\infty)$ not $[2,\infty)$ .

Then for obvious reasons its minimum value is does not exist .

Using calculus,

Let, $y=x+x^{-1}$

$\frac{dy}{dx}=1-x^{-2}$

For minima the derivative must equal to zero,

$1-x^{-2}=0$

$x=1,-1$

But in the question we are given $x≠1$ and $x>0$

Hence the minima doesn't exists.

Using the fact that 0.9999... is actually equal to 1, we can say that 0.000...0001 is equal to 0. As minimum value of X is 0.000...0001 and the value is 0, the answer is Does not Exist.