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anonymous
 5 years ago
Let \[x _{i}>0\] for i=1,2,3,...,n. For each positive integer k, prove that: \[(x _{i}^{k}+...+x _{n}^{k})/n≤(x _{i}^{k+1}+...x _{n}^{k+1})/(x_{1}+x_{2}+...+x_{n})\]
anonymous
 5 years ago
Let \[x _{i}>0\] for i=1,2,3,...,n. For each positive integer k, prove that: \[(x _{i}^{k}+...+x _{n}^{k})/n≤(x _{i}^{k+1}+...x _{n}^{k+1})/(x_{1}+x_{2}+...+x_{n})\]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Rewrite the equation as \[\frac{x_1+\cdots+x_n}{n} \le \frac{x_1^{k+1}+\cdots +x_n^{k+1}}{x_1^k +\cdots + x_n^k}\] Note that this is the same as \[\frac{x_1^1+\cdots + x_n^1}{x_1^0+\cdots +x_n^0}\le \frac{x_1^{k+1}+\cdots +x_n^{k+1}}{x_1^k +\cdots + x_n^k}\] There is a type of mean known as the Lehmer mean, which is \[L_a=\frac{x_1^a+\cdots +x^a_n}{x^{a1}_1+\cdots x^{a1}_n}\] Using a bit of calculus, namely taking the derivative with respect to a, you'll known that L_a is a monotone increasing function with respect to a. Because it is increasing, you have that \[p \le q \implies L_p \le L_q\] Note that your inequality is the same as \[L_1 \le L_k\] which (using Lehmer's inequality) is obviously true for all k greater than or equal to one, which encompasses all positive integers.
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