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Given 0 <= a <= b,

how do you prove that

sqrt(ab) <= (a+b)/2?

2007-09-03 08:04:55 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

0 <=(a-b)^2 by the definition of square
0<=a^2-2ab+b^2
4ab<=a^2+2ab+b^2 (adding 4ab to both sides)

So 4ab<=(a+b)^2

Now take square roots of both sides to get the answer.

2007-09-03 08:15:49 · answer #1 · answered by rothbach 2 · 1 0

Since 0 <= a <= b, sqrt(a) and sqrt(b) are well defined.
we know that any thing squared must be zero or positive:
0 <= (sqrt(a) - sqrt(b))^2 = a - 2sqrt(ab) + b
or: 2sqrt(ab) <= (a + b)
Half both sides and you get the answer!

2007-09-05 13:49:57 · answer #2 · answered by Hahaha 7 · 0 0

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