Let $$x _{i}0$$ for i=1,2,3,...,n. For each positive integer k, prove that: $$(x _{i}^{k}+...+x _{n}^{k})/n≤(x _{i}^{k+1}+...x _{n}^{k+1})/(x_{1}+x_{2}+...+x_{n})$$

Question

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 · Answer

Rewrite 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 $$L_a=\frac{x_1^a+\cdots +x^a_n}{x^{a-1}_1+\cdots x^{a-1}_n}$$ 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.