Question

Let \( x \text{ and } y \) be two positive real numbers. Then
\[ \left( x + \frac{1}{x} \right) \left( y + \frac{1}{y} \right) \] is greater than or equal to:
 

• A. 3
• B. 2
• C. 4
• D. 5
• E. 6

Hint:

The AM-GM inequality is useful in problems involving sums and products of terms. Remember that for positive real numbers \( a \) and \( b \), the inequality \( a + b \geq 2\sqrt{ab} \) holds, with equality when \( a = b \).

Answer 1

The Correct Option is C

We are given that \( x \) and \( y \) are positive real numbers, and we need to find the minimum value of the expression:
\[ \left( x + \frac{1}{x} \right) \left( y + \frac{1}{y} \right). \] Step 1: We apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to each factor individually:
\[ x + \frac{1}{x} \geq 2 \quad {and} \quad y + \frac{1}{y} \geq 2, \] because by the AM-GM inequality, the arithmetic mean of two positive numbers is always greater than or equal to their geometric mean, with equality holding when the numbers are equal.
Step 2: Now, multiply these two inequalities: \[ \left( x + \frac{1}{x} \right) \left( y + \frac{1}{y} \right) \geq 2 \times 2 = 4. \]
Thus, the minimum value of the expression \( \left( x + \frac{1}{x} \right) \left( y + \frac{1}{y} \right) \) is 4.
Therefore, the correct answer is option (C).