Simple inequality involving product of sums

Question

Suppose we have a sequence of numbers such that $0<x_1 \leq x_2 \leq x_3 \leq...\leq x_n$. Would it be possible to show that $(\sum \frac{1}{x_i})(\sum x_i) \geq n^2$?

Answer 1

You can use the Cauchy-Schwarz Inequality on the sequences $\left\{\dfrac1{\sqrt{x_1}},\dots,\dfrac1{\sqrt{x_n}}\right\}$ and $\{\sqrt{x_1},\dots,\sqrt{x_n}\}$ to have

$$ \left( \sum_{i=1}^n \left( \dfrac1{\sqrt{x_i}} \right)^2 \right) \left( \sum_{i=1}^n \sqrt{x_i}^2 \right) \ge \left(\sum_{i=1}^n \dfrac1{\sqrt{x_i}}\sqrt{x_i}\right)^2=\left(\sum_{i=1}^n 1\right)^2=n^2$$

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There are a number of ways to show this, of course, for instance using the AM-HM inequality on the $n$ positive numbers $x_1,x_2,\cdots,x_n$ as $$\text{A.M. of the $n$ numbers}=\dfrac{\sum_{i=1}^n x_i}n \ge \dfrac{n}{\sum_{i=1}^n \frac1{x_i}}=\text{H.M. of the $n$ numbers}$$

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To clarify the comments, you can also use a certain generalization of the Rearrangement inequality called the Chebyshev Inequality on the oppositely-sorted sequences $$x_n \ge x_{n-1} \ge \dots \ge x_1\\ \frac1{x_n} \le \frac1{x_{n-1}} \le \dots \le \frac1{x_1}$$ to get the desired result as well. In fact, the Chebyshev inequality can be proved using the Rearrangement Inequality (which is how Rearrangement inequality can be used to prove this without knowing Chebyshev's inequality, i.e. you would basically be doing the proof of Chebyshev's inequality given in the link above, if you scroll down in it a bit.)