$x_i > \frac{\sum_{j=1}^{n} x_j}{n} \leftrightarrow x_i > \frac{\sum_{j\not = i} x_j}{n-1}$

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?

• What is this $\frac{\sum_{i=i}^{n>=i} x_i}{n}$? – Math Lover May 29 '18 at 20:57
• @MathLover Poor notation. I was just trying to indicate that $x_i$ is included in the sum. I will change it. – user106860 May 29 '18 at 20:59

Note that $$x_i > \frac{\sum_{j=1}^{n} x_j}{n} \iff \color{red}{x_i} > \frac{\color{blue}{x_i} + \sum_{j\neq i} x_j}{n} \iff \frac{n\color{red}{x_i}-\color{blue}{x_i}}{n} > \frac{ \sum_{j\neq i} x_j}{n} \iff \cdots$$