Problem 5 - A Gentle Introduction to the Art of Mathematics

Question

Show that if \(x\) is a positive real number, then \(x+\frac{1}{x} \geq 2\).

Step-by-Step Solution

Answer

By applying the AM-GM Inequality and simplifying, we show that \( x + \frac{1}{x} \geq 2 \).

Step 1 - Define the given expression

Consider the expression provided: \[ f(x) = x + \frac{1}{x} \] where \( x \) is a positive real number.

Step 2 - Use the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality)

According to the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality), for any two positive numbers \( a \) and \( b \), we have that: \[ \frac{a + b}{2} \geq \sqrt{ab} \] To apply this, let \( a = x \) and \( b = \frac{1}{x} \).

Step 3 - Apply AM-GM Inequality

Substituting \( a = x \) and \( b = \frac{1}{x} \) into the AM-GM Inequality, we get: \[ \frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} \]

Step 4 - Simplify the inequality

Simplify the right side of the inequality: \[ \sqrt{x \cdot \frac{1}{x}} = \sqrt{1} = 1 \] Therefore, we have: \[ \frac{x + \frac{1}{x}}{2} \geq 1 \]

Step 5 - Final step

Multiply both sides of the inequality by 2 to isolate the given expression: \[ x + \frac{1}{x} \geq 2 \] Thus, we have proven that if \( x \) is a positive real number, then \( x + \frac{1}{x} \geq 2 \).

Key Concepts

inequality proof

The Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality) is a fundamental concept in algebra that provides a way to compare the arithmetic mean and geometric mean of positive numbers. The AM-GM Inequality states that for any set of positive real numbers, the arithmetic mean is always greater than or equal to the geometric mean.

Let's break down the inequality step by step:

• The Arithmetic Mean (AM) of two numbers \(a\) and \(b\) is given by: \( \frac{a + b}{2} \).
• The Geometric Mean (GM) of the same two numbers is: \( \sqrt{ab} \).

The AM-GM Inequality can be written as:
\[ \frac{a + b}{2} \ge\sqrt{ab} \] Applying this to the given problem, where the two numbers we are comparing are \(x\) and \(\frac{1}{x}\), we proceed to simplify it. This results in our desired inequality:
\[ x + \frac{1}{x} \ge 2 \] This proof elegantly shows why the Arithmetic Mean is always equal to or exceeds the Geometric Mean for positive real numbers.

positive real numbers

The concept of positive real numbers is essential to the proof above. A positive real number is any number greater than zero that is not imaginary or complex. Here, we deal with positive real numbers exclusively.

Positive real numbers have several properties:

• They are greater than zero: \(x > 0\).
• They are used in various mathematical disciplines, particularly in inequalities.
• They can be fractions, whole numbers, or decimals.

In our inequality, both \(x\) and \(\frac{1}{x}\) are positive real numbers. When working with inequalities or equations involving positive real numbers, ensuring that all values stay positive is crucial, as it guarantees valid operations such as taking square roots and geometric means.

This distinction helps avoid undefined or erroneous calculations that could arise from including zero or negative values.

algebraic manipulation

Algebraic manipulation involves rearranging and simplifying expressions to reveal underlying relationships and properties. In proving the inequality \(x + \frac{1}{x} \ge 2\), algebraic manipulation allows us to effectively apply the AM-GM Inequality.

Let's go through these steps:

• First, we start with the given function: \(f(x) = x + \frac{1}{x} \).
• We then apply the AM-GM Inequality to the values of \(a\) and \(b\): \(a = x\) and \(b = \frac{1}{x}\).
• So, the inequality becomes \( \frac{x + \frac{1}{x}}{2} \ge \sqrt{x \cdot \frac{1}{x}} \).
• Simplifying further, we get: \( \sqrt{x \cdot \frac{1}{x}} = \sqrt{1} = 1 \).
• Finally, rearranging and multiplying both sides by 2, we get our final expression: \( x + \frac{1}{x} \ge 2 \).

This step-by-step simplification illustrates that by careful algebraic manipulation, inequalities can be more easily understood and proven.

Such manipulations are powerful and frequently used tools in mathematics, from simple algebra to complex calculus, enabling us to break down and solve problems systematically.