Question

For two positive numbers \(a\) and \(b\), prove that $$ \sqrt{a b} \leq \frac{1}{2}(a+b) $$

Short answer

The inequality \( \sqrt{a b} \leq \frac{1}{2}(a+b) \) holds by expanding and simplifying into \( (a-b)^2 \geq 0 \).

Step-by-step solution

Understand the statement

The inequality we need to prove is \( \sqrt{a b} \leq \frac{1}{2}(a+b) \). This is known as the arithmetic mean-geometric mean inequality, a fundamental concept in algebra.

Set up the inequality

To prove \( \sqrt{a b} \leq \frac{1}{2}(a+b) \), it helps to rewrite it in a form that is easier to manipulate. Rearrange the inequality to show that multiplying both sides by 2 gives \( 2\sqrt{a b} \leq a + b \).

Square both sides

To simplify \( 2\sqrt{a b} \leq a + b \), square both sides to eliminate the square root: \( (2\sqrt{a b})^2 = 4ab \) and \( (a+b)^2 = a^2+2ab+b^2 \).

Compare the squared terms

The inequality \( 4ab \leq a^2 + 2ab + b^2 \) now needs to be shown to be true. Simplifying gives \( 0 \leq a^2 - 2ab + b^2 \).

Factor the quadratic expression

Notice that the expression \( a^2 - 2ab + b^2 \) can be factored as \( (a-b)^2 \). A square of a real number is always non-negative, \( (a-b)^2 \geq 0 \).

Conclude the inequality proof

Since \( (a-b)^2 \geq 0 \) is true for all real numbers, the squared-term comparison implies the original inequality \( 4ab \leq a^2 + 2ab + b^2 \) holds, thereby confirming that \( \sqrt{a b} \leq \frac{1}{2}(a+b) \).

Key concepts

Inequality Proof

The proof of the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality) is a clear demonstration of mathematical principles. This inequality states that, for two positive numbers \(a\) and \(b\), the geometric mean \(\sqrt{ab}\) is always less than or equal to the arithmetic mean \(\frac{1}{2}(a+b)\).

Proving such an inequality involves manipulating expressions to unveil underlying mathematical truths. We start by squaring both sides of the inequality. This is a crucial step as it removes the square root, making comparisons easier. First, ensure that both \(a\) and \(b\) are positive, since the geometric mean requires positive values.

Once squared, the inequality \((2\sqrt{ab})^2 \leq (a+b)^2\) transforms into a form that shows deeper relationships between \(a\) and \(b\). The squaring leads us to work with terms like \(4ab\) and \(a^2 + 2ab + b^2\), allowing easier algebraic manipulation to establish the proof.

Quadratic Expressions

Quadratic expressions play a key role in proving the AM-GM Inequality. Once we've squared the inequality \( 2\sqrt{ab} \leq a + b\), we encounter quadratic expressions in the form \(a^2 - 2ab + b^2\). These expressions are pivotal as they need to be factored, simplified, or compared to prove the inequality.

The expression \(a^2 - 2ab + b^2\) can be written in a neatly factored form as \((a-b)^2\). Recognizing and manipulating quadratics effectively requires practice but is immensely powerful. This squares a difference and shows a non-negative result, since any real number squared is \(\geq 0\).

The ability to turn a quadratic into a square is a profound algebraic technique that simplifies proving inequalities. It transforms the problem into checking whether a sum of squares is non-negative, which is an easier task mathematically.

Algebraic Manipulation

Algebraic manipulation is the heart of transforming and proving inequalities. It involves a series of well-thought-out steps to re-arrange and simplify expressions, making them easier to work with. For the AM-GM Inequality, we begin by re-casting the inequality into a form without square roots, which involves algebraic squaring.

Once squaring has transformed our inequality into a simpler form \((a-b)^2 \geq 0\), we need to manipulate it using algebraic rules. Recognizing that \((a-b)^2\) being a square is always non-negative completes the initial part of the proof.

Algebraic manipulation doesn't only involve rearranging terms; it demands precision and a methodical approach to work step-by-step. Remember always to check each transformation's validity, affirming that each side of an inequality remains true at each stage.