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Unformatted text preview: ANALYSIS EXERCISE 6–SOLUTIONS 1. a + 1 a 2 = ( a 1) 2 a ≥ . The equality holds iff a = 1. 2. Multiplying the inequalities a + b ≥ 2 √ ab, a + c ≥ 2 √ ac, c + b ≥ 2 √ cb, we obtain the desired. The equality holds iff a = b = c . 3. For all n ≥ 2 we have 1 n 2 < 1 n ( n 1) = 1 n 1 1 n . Hence 1 2 2 + 1 3 2 + ··· + 1 n 2 < 1 1 2 + 1 2 1 3 + ··· + 1 n 1 1 n = 1 1 n < 1 . 4. a) If x ≥ 0 then   x  =  x = x =  x  ; while if x < 0,   x  = x =  x  . If x ≥ 0 then  x  = x ≥ x ; while if x < 0 then  x  = x ≥ x . If x ≥ 0 then max { x, x } = x =  x  while if x < 0, max { x, x } = x =  x  . b) If x ≥ 0 and y ≥ 0 then xy ≥ 0 so  xy  = xy =  x  y  . If x ≥ 0 and y < 0 then xy ≤ 0 so  xy  = xy = x ( y ) =  x  y  . If x < 0 and y ≥ 0 then xy ≤ 0 so  xy  = xy = ( x ) y =  x  y  . If x < 0 and y < 0 then xy > 0 so  xy  = xy = ( x )( y ) =  x  y  . c) See lectures. d) By c),  x  =  ( x y ) + y  ≤  x y  +  y  ⇒  x    y  ≤  x y  . Similarly,  y    x  ≤  x y  . By considering the cases  x  ≥  y  and  x  <  y  we obtain  x   y  ≤  x y  . 1 2 ANALYSIS EXERCISE 6–SOLUTIONS 5. Base case : n = 2;  a 1 + a 2  ≤  a 1  +  a 2  , by the triangle inequality. The statement is true. Induction step: Suppose that  a 1 + a 2 + ··· + a n  ≤  a 1  +  a 2  + ··· +  a n  ....
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This note was uploaded on 02/25/2010 for the course MATHEMATIC 11007 taught by Professor Spyros during the Winter '10 term at Bristol Community College.
 Winter '10
 Spyros
 Inequalities

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