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I want to show that a number used in an average is larger than the average IFF it is larger than the average of the other numbers. That is, $$x_i > \frac{\sum_{j=1}^{n} x_j}{n} \leftrightarrow x_i > \frac{\sum_{j\not = i} x_j}{n-1}$$

For the $(\leftarrow)$ direction I have the following: $$ x_i > \frac{\sum_{j\not = i} x_j}{n-1} \implies \frac{n-1}{n}x_i > \frac{\sum_{j\not = i} x_j}{n} \implies \frac{n-1}{n}x_i + \frac{x_i}{n} > \frac{\sum_{j\not = i} x_j}{n}+\frac{x_i}{n} $$ which implies $$ x_i> \frac{\sum_{i=1}^n x_i}{n} $$

But I am unsure of the other direction. I think I can just reverse the steps? I.E. $$ x_i > \frac{\sum_{i=i}^{n} x_i}{n} \implies x_i \frac{n}{n-1} > \frac{\sum_{i=i}^{n} x_i}{n-1} \implies x_i > \frac{\sum_{j\not= i} x_j}{n-1} $$ where the last inequality follows from subtracting $\frac{x}{n-1}$

Is this correct?

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  • $\begingroup$ What is this $\frac{\sum_{i=i}^{n>=i} x_i}{n} $? $\endgroup$ – Math Lover May 29 '18 at 20:57
  • $\begingroup$ @MathLover Poor notation. I was just trying to indicate that $x_i$ is included in the sum. I will change it. $\endgroup$ – user106860 May 29 '18 at 20:59
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Note that $$x_i > \frac{\sum_{j=1}^{n} x_j}{n} \iff \color{red}{x_i} > \frac{\color{blue}{x_i} + \sum_{j\neq i} x_j}{n} \iff \frac{n\color{red}{x_i}-\color{blue}{x_i}}{n} > \frac{ \sum_{j\neq i} x_j}{n} \iff \cdots$$

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  • $\begingroup$ I see. Thank you! $\endgroup$ – user106860 May 29 '18 at 21:02

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