Question

If \(x\) and \(y\) are both positive, then the minimum value of \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right)\) is (a) 0 (b) 1 (c) 2 (d) 4

Short answer

Question: Find the minimum value of $(x+y)\left(\frac{1}{x}+\frac{1}{y}\right)$ where $x$ and $y$ are positive. Solution: The minimum value of $(x+y)\left(\frac{1}{x}+\frac{1}{y}\right)$ is 2 when x = y.

Step-by-step solution

Rewrite the given expression

We first rewrite the given expression as a sum, like this: \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right) = \frac{x}{y} + \frac{y}{x} +2\)

Apply AM-GM inequality

By applying the AM-GM inequality to the terms \(\frac{x}{y}\) and \(\frac{y}{x}\), we get: \(\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right) \geq \sqrt{\frac{x}{y}\cdot \frac{y}{x}}\) Simplifying the inequality, we have: \(\frac{x^2+y^2}{2xy} \geq 1\)

Rewrite the inequality

Now we multiply through by \(2xy\) to get rid of the fractions: \(x^2 + y^2 \geq 2xy\) Now, we add \(2\) on both sides: \(x^2 + y^2 +2 \geq 2xy + 2\) Recall the rewritten expression of our problem: \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right) = \frac{x}{y} + \frac{y}{x} +2\) From the inequality above, we see that the minimum value for our expression can be found when \(x^2+y^2 = 2xy\).

Find the minimum value

So, the minimum value for our expression is when: \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right) = 2xy + 2\) But we want to find the answer from the multiple-choice options. Plug in the answer choices (a, b, c, d) and find which one is true when \(x=y\): (a) 0: \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right) = 0\) This is not true when \(x=y\), because both \(x\) and \(y\) are positive. (b) 1: \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right) = 1\) This is not true when \(x=y\). (c) 2: \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right) = 2\) This is true when \(x=y\), because \((2x)\left(\frac{2}{x}\right) = 2\). (d) 4: \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right) = 4\) This is not true when \(x=y\). Thus, we see that the minimum value of \((x+y)\left(\frac{1}{x}+\frac{1}{y}\right)\) is 2, which corresponds to option (c).

Key concepts

Minimum Value

Understanding the concept of minimum value in mathematical expressions is crucial for solving optimization problems. In our exercise, the task is to find the minimum possible value of the expression \( (x+y)\left(\frac{1}{x}+\frac{1}{y}\right) \), given that \(x\) and \(y\) are both positive numbers. It's important to note that 'minimum value' refers to the smallest value that this expression can ever reach, no matter the positive values of \(x\) and \(y\).

When working with minimum value problems, it's often helpful to visualize or conceptualize how the expression behaves as the variables change. In terms of algebra, finding the minimum value can sometimes involve the use of inequalities and sometimes calculus, depending on the complexity of the function. For the expression in our problem, we employed the AM-GM inequality, a well-known method in algebra that relates the arithmetic mean and geometric mean, to determine when the expression reaches its smallest value. This clever use of AM-GM inequality simplifies the problem and avoids unnecessary complex calculations.

Inequality Problem Solving

In inequality problem solving, one of the most useful tools is the AM-GM inequality, which states that for any non-negative numbers, the arithmetic mean is always greater than or equal to the geometric mean. This can be written as \( \frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot \ldots \cdot a_n} \).

In our example, we use this inequality to find when the sum \( \frac{x}{y} + \frac{y}{x} \) is at its minimum, knowing that \(x\) and \(y\) are positive. After applying AM-GM, we unravel the inequality and relate it back to our original expression to conclude the minimum value. This clear and strategic approach underscores the significance of understanding and applying inequalities to crack mathematics problems, especially those that seem convoluted at first glance. It is an essential skill that, once mastered, can make algebraic challenges much more manageable and can be expanded to tackle a wide range of problems, leading to correct and quick solutions.

Algebraic Expressions

In our exercise, we deal with an algebraic expression that contains variables \(x\) and \(y\), and operations including addition, division, and multiplication. Algebraic expressions are the backbone of algebra; they represent mathematical relationships using symbols. The beauty of using algebraic expressions is that they can describe patterns, generalize problems, and allow us to manipulate mathematical relationships to find desired results.

Through the given situation, rewriting expressions and applying algebraic rules help us uncover deeper relations or simplify complex problems. For instance, the step-by-step solution began by expanding the given expression to inclusde \(\frac{x}{y} + \frac{y}{x} + 2\), enabling us to apply the AM-GM inequality more effectively. This manipulation reflects a core component of algebra: reformatting expressions in ways that reveal hidden pathways to solutions. It is through these expressions that students can learn to solve a wide range of problems and understand why particular strategies, such as applying inequalities, are effective in certain contexts.