Prove the Inequality a/b + b/a ≥ 2 for a, b ∈ ℝ⁺

Question

Let a, b be real numbers in ℝ⁺ (that is, a > 0 and b > 0). Prove the inequality:

a/b + b/a ≥ 2

Answer 1

For any real number x, x² ≥ 0.

So in this case, (a - b)² is the square of the real number a - b.

Therefore, (a - b)² ≥ 0.

Since (a - b)² ≥ 0,

a² - 2ab + b² ≥ 0 ⇔ a² + b² ≥ 2ab ⇔ (a² + b²) / ab ≥ 2 ⇔ (a / b) + (b / a) ≥ 2.

Equality occurs if and only if a − b = 0, i.e. a = b.