Kerala KEAM 5 June 2024 Solved Paper

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).Quick Tip: 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 $a = b$.