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Prove this statement: For all real numbers x and y, x^2+y^2 ≥2xy.

a) True
b) False

1 Answer

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Final answer:

The statement that for all real numbers x and y, x^2 + y^2 is greater than or equal to 2xy is true. It is derived from the algebraic identity (x - y)^2 = x^2 - 2xy + y^2, and since squares are non-negative, this proves the inequality.

Step-by-step explanation:

The statement that for all real numbers x and y, x^2 + y^2 ≥ 2xy is true. We can prove this using the concept of squares of real numbers being non-negative. The statement is essentially a rephrasing of the algebraic identity (x - y)^2 = x^2 - 2xy + y^2, which is always non-negative because it represents the square of the difference between x and y. Since (x - y)^2 is non-negative, it follows that x^2 - 2xy + y^2 ≥ 0. Rearranging this inequality, we get x^2 + y^2 ≥ 2xy, confirming the initial statement as true.

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