\( ( x + 2 y ) ( \frac { 1 } { x } + \frac { 1 } { 2 y } ) \geq 4 \)

To solve the inequality \[ ( x + 2y ) \left( \frac{1}{x} + \frac{1}{2y} \right) \geq 4, \] we can start by simplifying the left-hand side. We can distribute the terms: \[ ( x + 2y ) \left( \frac{1}{x} + \frac{1}{2y} \right) = (x + 2y) \cdot \frac{1}{x} + (x + 2y) \cdot \frac{1}{2y}. \] Calculating each term separately: 1. The first term: \[ (x + 2y) \cdot \frac{1}{x} = 1 + \frac{2y}{x}. \] 2. The second term: \[ (x + 2y) \cdot \frac{1}{2y} = \frac{x}{2y} + 1. \] Now, combining these results, we have: \[ 1 + \frac{2y}{x} + \frac{x}{2y} + 1 = 2 + \frac{2y}{x} + \frac{x}{2y}. \] Thus, the inequality becomes: \[ 2 + \frac{2y}{x} + \frac{x}{2y} \geq 4. \] Subtracting 2 from both sides gives: \[ \frac{2y}{x} + \frac{x}{2y} \geq 2. \] Now, we can apply the AM-GM inequality (Arithmetic Mean - Geometric Mean inequality) to the terms \(\frac{2y}{x}\) and \(\frac{x}{2y}\): \[ \frac{2y}{x} + \frac{x}{2y} \geq 2 \sqrt{\frac{2y}{x} \cdot \frac{x}{2y}} = 2 \sqrt{1} = 2. \] This shows that the inequality \[ \frac{2y}{x} + \frac{x}{2y} \geq 2 \] holds true for all positive \(x\) and \(y\). Therefore, the original inequality \[ ( x + 2y ) \left( \frac{1}{x} + \frac{1}{2y} \right) \geq 4 \] is satisfied for all positive \(x\) and \(y\). Thus, the solution to the inequality is: \[ \text{The inequality holds for all } x > 0 \text{ and } y > 0. \]