# exercises_6 - Analysis 1 Exercises 6 1 Prove that(a > 0 a 1...

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Analysis 1: Exercises 6 1. Prove that ( a > 0) a + 1 a 2 . 2. Prove that ( a, b, c 0) ( a + b )( a + c )( b + c ) 8 abc . 3. Prove that ( n N ) ( n 2) 1 2 2 + 1 3 2 + · · · + 1 n 2 < 1 . 4. Let x, y R . Show that a) | - x | = | x | , x 6 | x | , | x | = max { x, - x } ; b) | xy | = | x | | y | ; c) | x + y | 6 | x | + | y | ; d) || x | - | y || 6 | x - y | . 5. Prove that ( n N + )( a 1 , a 2 , . . . , a n R )[ | a 1 + a 2 + · · · + a n | ≤ | a 1 | + | a 2 | + · · · + | a n | ] . 6. Prove the Cauchy-Bunyakovski-Schwarz inequality: ( n N + )( a 1 , a 2 , . . . , a n R ) ( b 1 , b 2 , . . . , b n R ) n X k =1 a k b k 2 n X k =1 a 2 k n X k =1 b 2 k . Hint : For all x R , n X k =1 ( a k x + b k ) 2 = x 2 n X k =1 a 2 k + 2 x n X k =1 a k b k + n X k =1 b 2 k 0 . 7. Let a 1 , a 2 , . . . , a n be positive real numbers. Their arithmetic mean A n and harmonic mean H n are defined by A n = a 1 + a 2 + . . . + a n n , H - 1 n = 1 n 1 a n + 1 a 2 + . . . + 1 a n Deduce from the Cauchy-Bunyakovski-Schwarz inequality that H n 6 A n . 8. Prove that ( n N + )( a, b R + ) a + b 2 n a n + b n 2 . 9. The Cauchy inequality . For any n non-negative numbers a 1 , a 2 . . . , a n the inequality 1 n n X k =1 a k > n v u u t n Y k =1 a

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