Question

Proving algebraic statements with direct proofs. Prove each of the following statements using a direct proof.

For any positive real numbers,x and y, x + y ≥ √xy

Answer 1

The proof for the inequality x + y ≥ √xy for positive real numbers involves the algebraic manipulation showing that the square of the sum is at least as large as twice the product and that the arithmetic mean is greater than or equal to the geometric mean.

Step-by-step explanation:

The question asks us to prove that for any positive real numbers, x and y, the inequality x + y ≥ √xy holds. A direct proof can be approached by manipulating the inequality and using algebraic properties.

Proof

Let's start by considering the square of the sum of two positive real numbers:

1. (x + y)² = x² + 2xy + y².
2. Since the square of a real number is nonnegative, the right side of the equation is at least as large as 2xy. Hence, x² + 2xy + y² ≥ 2xy.
3. Dividing the entire inequality by 2 yields, (x² + 2xy + y²)/2 ≥ xy.
4. Applying the principle that the arithmetic mean is greater than or equal to the geometric mean for positive numbers, we have (x/2 + y/2) ≥ √xy.
5. Simplifying further, x + y ≥ 2√xy, which leads to the final result x + y ≥ √xy after dividing by 2 under the square root on the right side.

This completes the proof that x + y ≥ √xy for any positive real numbers x and y.