Show that:$\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}} \geq n^2 $

Question

Show that:$$\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}} \geq n^2 $$ The following hints are also given: $$\left(\frac{x}{y} + \frac{y}{x} \geq 2 \right) \land x,y \gt 0$$

Base Case: For n = 2

$$\left(1+2\right) \cdot \left(\frac{1}{1}+\frac{1}{2}\right) \geq 2^2$$

Inductive hypothesis:

$$\sum_{i=1}^{n+1} x_{i} \cdot \sum_{i=1}^{n+1} \frac{1}{x_{i}} \geq \left(n+1\right)^2 = n^2+2n+1$$

Inductive step:

$$\sum_{i=1}^{n+1} x_{i} \cdot \sum_{i=1}^{n+1} \frac{1}{x_{i}} = \left(\sum_{i=1}^n x_{i}+(n+1)\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_{i}}+\frac{1}{n+1}\right) = \left(\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) + \left((n+1) \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) + \left(\sum_{i=1}^n x_{i} \cdot \frac{1}{n+1}\right) + (n+1) \cdot \frac{1}{n+1}$$

Final words:

I came to the conclusion that:

$$\left(\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) = n^2$$

I inserted for n = 1 so that $$\left((n+1) \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) = \frac{2}{1} \land \left(\sum_{i=1}^n x_{i} \cdot \frac{1}{n+1}\right) = \frac{1}{2}$$ Since $$\left(\frac{x}{y} + \frac{y}{x} \geq 2 \right) \land x,y$$ was given as a hint in the beginning I thougt I can say that $$\left((n+1) \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) + \left(\sum_{i=1}^n x_{i} \cdot \frac{1}{n+1}\right) = 2n$$

Furthermore it's obvious that: $$(n+1) \cdot \frac{1}{n+1} = 1$$

It's pretty standard proof by induction and I hope you can maybe give me some advices on what I could have done differently and verify the legitimacy of this proof.

Answer 1

Alternative way : It's a direct consequence of the inequality of Chebyshev .

Answer 2

One must assume $\,x_i \gt 0\,$ otherwise the inequality doesn't necessarily hold true.

Once assumed that all $\,x_i\,$ are positive, this is just the AM $\ge$ HM mean inequality:

$$\frac{\sum_{i=1}^n x_{i}}{n} \;\ge\; \frac{n}{\sum_{i=1}^n \frac{1}{x_{i}}}$$