<s>[INST] Show that:$\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}} \geq n^2 $ 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.
 [/INST] Alternative way :
It's a direct consequence of the inequality of Chebyshev .
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