2y² / x + y >= x + y.

This is a mathematical inequality: \( \frac{2y^2}{x} + y \geq x + y \). To solve this, we can simplify it:

1. Subtract y from both sides: \( \frac{2y^2}{x} \geq x \)
2. Since we are given that \( y > 0 \), and the context implies \( x > 0 \) (otherwise the division by x would be undefined or the inequality direction would change), we can multiply both sides by \( x \) without changing the inequality sign: \( 2y^2 \geq x^2 \)
3. Taking the square root of both sides (since x and y are implied to be positive): \( \sqrt{2}y \geq x \)

Answer: The inequality simplifies to \( \sqrt{2}y \geq x \) given the conditions \( y > 0 \) and \( x > 0 \).