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Part of 2003 Putnam

Problems(2)

Putnam 2003 A2

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6/22/2011
Let a1,a2,⋯ ,ana_1, a_2, \cdots , a_n and b1,b2,⋯ ,bnb_1, b_2,\cdots, b_n be nonnegative real numbers. Show that (a1a2⋯an)1/n+(b1b2⋯bn)1/n≤((a1+b1)(a2+b2)⋯(an+bn))1/n(a_1a_2 \cdots a_n)^{1/n}+ (b_1b_2 \cdots b_n)^{1/n} \le ((a_1 + b_1)(a_2 + b_2) \cdots (a_n + b_n))^{1/n}
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Putnam 2003 B2

Source:

6/23/2011
Let nn be a positive integer. Starting with the sequence 1,12,13,⋯ ,1n1,\frac{1}{2}, \frac{1}{3} , \cdots , \frac{1}{n}, form a new sequence of n−1n -1 entries 34,512,⋯ ,2n−12n(n−1)\frac{3}{4}, \frac{5}{12},\cdots ,\frac{2n -1}{2n(n -1)}, by taking the averages of two consecutive entries in the first sequence. Repeat the averaging of neighbors on the second sequence to obtain a third sequence of n−2n -2 entries and continue until the final sequence consists of a single number xnx_n. Show that xn<2nx_n < \frac{2}{n}.
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