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# inequality watch

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1. if and

prove

2. a^3/[a^2 + b^2] = a - [ab^2]/a^2 + b^2].
Do likewise for the other two expressions.
You then have 1 - {[ab^2]/[a^2 + b^2] + .. + ..}
Since arithmetic mean is greater than or equal to geometric mean it follows that [a^2 + b^2]/2 >=ab so that 1/[a^2 + b^2] <=1[2ab].
Similarly for the other two expressions.
Result follows.

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Updated: March 8, 2008
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