# Answer to Question #14849 in Real Analysis for sumaira

Question #14849

prove that [1/2(a+b)]^2<=1/2(a^2+b^2)

Expert's answer

Here is the proof:

0 <= (a-b)²

0 <= a²-2ab+b²

2ab <= a²+b² ==>

1/2ab <= 1/4(a²+b²) ==>

1/4(a²+2ab+b²) <= 1/2(a²+b²) ==>

[1/2(a+b)]² <= 1/2(a²+b²).

0 <= (a-b)²

0 <= a²-2ab+b²

2ab <= a²+b² ==>

1/2ab <= 1/4(a²+b²) ==>

1/4(a²+2ab+b²) <= 1/2(a²+b²) ==>

[1/2(a+b)]² <= 1/2(a²+b²).

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