Answer to Question #14849 in Real Analysis for sumaira

Question

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

Answer 1

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).