Answer

Let’s assume further that x is positive real.

We start with a simple fact that ANY squares are non-negative. The idea is we “choose” a square intelligently to suit our purpose. Also not that x does NOT equal zero.

By experience,
I choose the square ( x^(1/2) – 1 / [ x^(1/2)] )^2.
It must be non-negative for whatever positive real values of x.
Thus,

( x^(1/2) – 1 / [ x^(1/2)] )^2 >= 0

Expanding and regrouping, we get,

x – 2 + 1/x >= 0

or equivalently,

x + 1/x >= 2

Thus, minimum is 2.
With equality when x^(1/2) = 1 / [ x^(1/2)], or
x = 1.

Done.

Alternative Method.

One can use two variable AM – GM,
For any two positive reals m and n, we have

( m + n )/ 2 >= (mn)^(1/2)

With equality if and only if m = n.

Just set m = x and n = 1/x, to get,

( x + 1/x )/ 2 >= 1

or x + 1/x >= 2

Again, the minimum values is 2 with equality if and only if x = 1 .