Question

Show that the sum of any positive number and its reciprocal cannot be less than 2 .

Short answer

Let x be a positive number and its reciprocal be 1/x. To show that their sum cannot be less than 2, we will manipulate the inequality x + 1/x ≥ 2. First, multiply both sides by x, giving us \(x^2 + 1 ≥ 2x\). Rearrange the inequality into a quadratic form: \(x^2 - 2x + 1 ≥ 0\). Factoring, we find \((x - 1)^2 ≥ 0\). Since this is always true, the sum of any positive number and its reciprocal cannot be less than 2.

Step-by-step solution

Write the inequality

Let x be a positive number and let its reciprocal be 1/x. We need to show that their sum, x + 1/x, cannot be less than 2, or mathematically, x + 1/x ≥ 2.

Multiply both sides by x

As the inequality contains a fraction (1/x), it is easier to deal with if we multiply both sides by x. Note that since x is a positive number, multiplying both sides by x does not change the direction of the inequality. We will have: \(x(x + \frac{1}{x}) ≥ 2x\) This simplifies to: \(x^2 + 1 ≥ 2x\)

Rearrange the inequality

We arrange the inequality in a quadratic inequality form by moving all the terms to one side. This gives us: \(x^2 - 2x + 1 ≥ 0\)

Study the quadratic function

The left side of our inequality looks like a quadratic function. Factor the expression to find a perfect square: \((x - 1)^2 ≥ 0\)

Interpret the meaning of the inequality

The inequality we have now, \((x - 1)^2 ≥ 0\), indicates that the square of a real number (x - 1) cannot be negative. As this statement is always true, it means that for any positive number x, and its reciprocal 1/x, their sum cannot be less than 2. Therefore, we have proven the original inequality. x + 1/x ≥ 2

Key concepts

Quadratic Functions

Quadratic functions are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These functions create a parabola when graphed on a coordinate plane. Understanding their components and how they behave is key to solving inequalities such as the one in the exercise.

In the step-by-step solution, we transformed the inequality \(x + \frac{1}{x} \geq 2\) into a quadratic inequality \(x^2 - 2x + 1 \geq 0\). This expression was then factorized into a perfect square, \((x-1)^2\), which is a special type of quadratic function.

Quadratic functions can be solved by factoring, using the quadratic formula, or completing the square. In this exercise, recognizing \((x - 1)^2\) as a perfect square was crucial to proving that the inequality holds for any positive \(x\). By rearranging and factoring, we demonstrated that the expression on the left side of the inequality is always non-negative, thus confirming the solution.

Reciprocals

A reciprocal of a number \(x\) is simply \(\frac{1}{x}\). If you multiply a number by its reciprocal, you get 1: \(x \cdot \frac{1}{x} = 1\). This property is fundamental in understanding how numbers and their reciprocals behave when combined in equations or inequalities.

In our exercise, we examined the sum of a positive number \(x\) and its reciprocal, \(\frac{1}{x}\). We needed to prove that this sum can never be less than 2. The manipulation involved multiplying both sides of the starting inequality by \(x\) to eliminate the fraction, simplifying our calculations.

Understanding how reciprocals interact with other mathematical operations helps in solving a wide range of problems. When dealing with positive numbers, reciprocals ensure we are only juggling values that make logical sense and adhere to the inequality rules we seek to prove.

Perfect Square

A perfect square is an expression of the form \((x - h)^2\) or \(x^2 - 2hx + h^2\), where \(h\) is a constant. Perfect squares are crucial in algebra because they are always non-negative. They reach their minimum value of zero at \(x = h\).

In this exercise, we turned the quadratic inequality \(x^2 - 2x + 1 \geq 0\) into a perfect square \((x - 1)^2 \geq 0\). Recognizing a perfect square simplifies proving the inequality, as it reveals the minimum value: zero.

• For \((x - 1)^2 \), the expression equals zero when \(x = 1\).
• Since squares are always non-negative, \((x - 1)^2\) cannot be less than zero.

This analysis confirms that the sum of a number and its reciprocal \(x + \frac{1}{x}\) has the minimum result of 2. Thus, using perfect squares helps effortlessly resolve the inequality we started with.