The correct option is (D) 4.
A.M ≥ H.M
\(\frac{x + y}{2}\) ≥ \(\frac{2xy}{x+y}\) → \(\frac{2}{\frac{1}{x} + \frac{1}{y}}\)
= \(\frac{x + y}{2}\) ≥ \(\frac{2}{\frac{1}{x} + \frac{1}{y}}\)
= (x + y) (\({\frac{1}{x} + \frac{1}{y}}\)) ≥ 4
= (\({\frac{1}{x} + \frac{1}{y}}\)) ≥ 4 (x + y = 1)
Minimum Value = 4.