Key idea. Subtract a likely lower bound and express the result as a square.

1. For \(x>0\),

   \[x+\frac1x-2=\frac{x^2-2x+1}{x}=\frac{(x-1)^2}{x}\ge0.\]

2. The numerator is non-negative and the denominator is positive. Thus \(x+1/x\ge2\).

3. Equality holds when \((x-1)^2=0\), namely at \(x=1\). This allowed value gives 2, so the lower bound is attained.

Remember. A minimum proof needs a bound and a value that attains it.