Final answer:
To prove that for any positive real numbers x and y, x + y ≥ √xy, one can square both sides of the inequality to get rid of the square root and demonstrate that the left side is indeed greater or equal to the right side.
Step-by-step explanation:
The correct answer is option (a), which involves proving for any positive real numbers, x and y, that x + y ≥ √xy. To prove this, let's use the idea that when two positive numbers multiply, the answer has a positive sign. By squaring both sides of x + y ≥ √xy, we aim to remove the square root and compare the squares of both sides.
Firstly, (x + y)^2 ≥ ( √xy )^2 simplifies to x^2 + 2xy + y^2 ≥ xy. Now, we can see that x^2 and y^2 are certainly positive since they are squares of positive numbers, and 2xy is also positive. In this case, it's clear that x^2 + 2xy + y^2 is greater than or equal to xy, which verifies that x + y ≥ √xy.
To prove that for any positive real numbers x and y, x + y ≥ √xy, we can start by squaring both sides of the inequality. This gives us (x + y)² ≥ xy. Expanding the left side of the inequality, we have x² + 2xy + y² ≥ xy. Rearranging the terms, we get x² + y² ≥ xy - 2xy, which simplifies to x² + y² ≥ -xy. Since x, y, and xy are all positive, -xy is negative. Therefore, x² + y² ≥ -xy is true for any positive real numbers x and y, which proves the original statement.