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1.Given that a+b=1,prove a^2+b^2 > ab 2.Given a+b=1,prove a+b > ab explain both of the briefly.

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For question 1, my original thinking is method II, but in view of question 2, I've found the better, and more straightfoward, method I. 1. Method I For all real values of a and b, a^2 + b^2 - ab = a^2 + (1 - a)^2 - a(1 - a) = a^2 + (1 - 2a + a^2) + (a^2 - a) = 3a^2 - 3a + 1 = 3(a^2 - a) + 1 = 3[(a - 1/2)^2 - 1/4] + 1 = 3(a - 1/2)^2 + 1/4 > 0 for all real values of a Therefore, a^2 + b^2 > ab Method II For all real values of a and b, a^2 + b^2 - ab = (a + b)^2 - 3ab = 1 - 3ab To prove a^2 + b^2 > ab, i.e. a^2 + b^2 - ab > 0, we have to prove 1 - 3ab > 0, i.e. 3ab < 1. Now, 3ab < 1 iff ab < 1/3 iff a(1 - a) < 1/3 iff a^2 - a + 1/3 > 0 iff (a - 1/2)^2 + 1/12 > 0 which is true for all real values of a Therefore, a^2 + b^2 > ab 2. For all real values of a and b, a + b - ab = a(1 - b) + b = a^2 + b (1 - b = a) = a^2 - a +1 (b = 1 - a) = (a - 1/2)^2 + 3/4 > 0 Therefore, a + b > ab 2008-04-28 23:37:33 補充： 如果有亂碼...> 係「大過」符號

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