x + 1/x ≥ 2
⇒ \(x\) + \(\frac{1}{x}\) - 2 ≥ 0
⇒ \(\frac{x^2-2x+1}{x}\) ≥ 0
Case I : x2 – 2x + 1 ≥ 0 and x > 0
(x – 1)2 ≥ 0 and x > 0
So, by taking intersection x > 0
Case II : x2 – 2x + 1 ≤ 0 and x < 0
(x – 1)2 ≤ 0 and x < 0
Square term is always positive so case II is irrelevant.
Then,
The final answer of question is x ∈ (0, ∞).