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Unformatted text preview: Solution to HW1 September 16, 2011 1.49 x 2 + y 2 + z 2 xy yz zx = 1 2 [( x y ) 2 + ( y z ) 2 + ( z x ) 2 ] ≥ which implies that x 2 + y 2 + z 2 ≥ xy + yz + zx. 1.52 For all real a 6 = 0 , a + 1 a 2 =  a  + 1  a  2 = a 2 + 1 2  a   a  = (  a   1) 2  a  ≥ , which implies  a + 1 /a  ≥ 2 . 1.54 Put x = 3 √ a 1 ,y = 3 √ a 2 ,z = 3 √ a 3 and what we need to prove becomes x 3 + y 3 + z 3 3 ≥ xyz. This is true because x 3 + y 3 + z 3 3 xyz = 1 3 ( x 3 + y 3 + z 3 3 xyz ) = 1 3 ( x + y + z )( x 2 + y 2 + z 2 xy yz zx ) ≥ . Here we used the fact that a 1 ,a 2 ,a 3 are positive and the conclusion proved in 1.49. 1 1.84 For n=1, the equation holds because LHS = 1 RHS = 1 2 = 1 . Assume that when n = k we already have 1 + 3 + ··· + (2 k 1) = k 2 . When n = k + 1 , LHS = 1 + 3 + ··· + (2 k 1) + [2( k + 1) 1] = k 2 + 2 k + 1 = ( k + 1) 2 = RHS, which means that the equation also holds. Therefore it is true for all positive integers. 1.85 For n = 1 , we have LHS = 1 1 · 3 = 1 3 = 1 2 · 1 + 1 = RHS. Assume that when n = k we already have 1 1 · 3 + 1 3 · 5 + ··· + 1 (2 k 1)(2 k + 1) = k 2 k + 1 , when n = k + 1 , LHS = 1 1 · 3 + 1 3 · 5 + ··· + 1 (2 k 1)(2 k + 1) + 1 (2 k + 1)(2 k + 3) = k 2 k + 1 + 1 (2 k + 1)(2 k + 3) = k (2 k + 3) + 1 (2 k + 1)(2 k + 3) = (2 k + 1)( k + 1) (2 k + 1)(2 k + 3) = k + 1 2 k + 3 = RHS, which means that the equation also holds. Therefore it is true for all positive integers. 1.89 For n = 1 , we have LHS = 1 3 = 1 = 1 4 1 2 · 2 2 = RHS. 2 Assume that when n = k we already have 1 3 + 2 3 + ··· + k 3 = 1 4 k 2 ( k + 1) 2 , when n = k + 1 , LHS = 1 3 + 2 3 + ··· + k 3 + ( k + 1) 3 = 1 4 k 2 ( k + 1) 2 + ( k + 1) 3 = 1 4 ( k + 1) 2 ( k 2 + 4( k + 1)) = 1 4 ( k + 1) 2 ( k + 2) 2 = RHS, which means that the equation also holds. Therefore it is true for all positive integers....
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This note was uploaded on 02/28/2012 for the course AMS 510 taught by Professor Feinberg,e during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Feinberg,E

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