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Unformatted text preview: k a k 22 k a k 2 + k a k 4 k b k 2 ( a b ) 2 = k a k 21 + k a k 2 k b k 2 ( a b ) 2 ! . 1 Since a b 6 = 0, a 6 = 0, hence k a k > 0. Therefore, the above can be rearranged to yield ( a b ) 2 k a k 2 k b k 2 . Taking the square root yields the result. From the proof, it is clear that the CauchSchwartz inequality is, in fact, strict unless for some , a = b , meaning that a and b are collinear. CauchySchwartz, in turn, implies the following. Theorem 2. For any a,b R N , k a + b k k a k + k b k . Proof k a + b k 2 = ( a + b ) ( a + b ) = k a k 2 + 2 a b k b k 2 k a k 2 + 2  a b  + k b k 2 k a k 2 + 2 k a kk b k + k b k 2 = ( k a k + k b k ) 2 . The second inequality follows form Theorem 1. Taking square roots yields the result. 2...
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 Fall '08
 JohnNachbar
 Economics

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