<p>If \(x, y \in R^+\) such that \(x + y = 8\), then find the minimum value of \(\left(1 + \dfrac{1}{x}\right)\left(1 + \dfrac{1}{y}\right)\).</p>
Step-by-Step Solution
Key Concept: Expand the product and use AM-GM inequality on the resulting terms, or recognize this as a constrained optimization where the minimum occurs when variables are equal by symmetry and convexity.
<p><strong>Step 1:</strong> Expand the product:</p><p>$$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) = 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy}$$</p><p><strong>Step 2:</strong> Since $x + y = 8$, use the constraint to rewrite $\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = \frac{8}{xy}$</p><p>$$\text{Expression} = 1 + \frac{8}{xy} + \frac{1}{xy} = 1 + \frac{9}{xy}$$</p><p><strong>Step 3:</strong> To minimize this, we need to maximize $xy$. By AM-GM inequality:</p><p>$$\frac{x+y}{2} \geq \sqrt{xy}$$</p><p>$$\frac{8}{2} \geq \sqrt{xy} \implies 4 \geq \sqrt{xy} \implies xy \leq 16$$</p><p>Equality holds when $x = y = 4$.</p><p><strong>Step 4:</strong> Maximum value of $xy$ is $16$, so minimum value of the expression is:</p><p>$$1 + \frac{9}{16} = \frac{25}{16}$$</p><p>∴ Answer: $\dfrac{25}{16}$</p>
Correct Answer: \(\dfrac{25}{16}\)