Final answer:
By assuming the contrary, we identify a contradiction using the derivative of the function f(x) = x + 1/x, which shows that the equation x + 1/x is always greater than or equal to 2 for any positive real number x.
Step-by-step explanation:
To prove by contradiction that for any positive number x in the set of positive real numbers, where x is not equal to 0, the equation x + 1/x ≥ 2 holds true, let's start by assuming the opposite, that is, there exists some x > 0 for which x + 1/x < 2.
Let's consider the function f(x) = x + 1/x. For this function, the critical points occur where f'(x) = 0. Calculating f'(x), we get f'(x) = 1 - 1/x^2. Setting this equal to zero gives us x = 1 as the critical point. Evaluating f(x) at x = 1, we find that f(1) = 1 + 1/1 = 2. Now, since f(x) is decreasing for x < 1 and increasing for x > 1, and we know that x + 1/x ≥ 2 at x = 1, it follows that f(x) ≥ 2 for all x > 0. This contradicts our initial assumption, hence proving that for all positive x, the equation x + 1/x ≥ 2 is true.