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With groups of sizes $x_1, x_2, \dots, x_n$ adding to $G$, the average size is _____. The chance of an individual belonging to group 1 is _____. The expected size of his or her group is $E(x) = x_1 (x_1/G)+\dots+x_n(x_n/G)$. *Prove $\sum_1^n x_i^2/G \ge G/n$.

The first blank should be $\frac Gn$. The second blank should be $\frac {x_1}G$. In the last statement of the question, I see the sum of squares and the square of sum. In common sense, the square of sum must be greater than or equal to the sum of squares because of extra terms in the former. Therefore, $G^2\ge\sum_{i=1}^nx_i^2$. But it seems that the sign reverses because of division by $n$. How do we prove that?

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$(\displaystyle\sum_{i}^{n}x_{i})^{2}\leq (\sum_{i}^{n}x_{i}^{2})(\sum_{i}^{n}1)=n(\sum_{i}^{n}x_{i}^{2}) \implies G/n \leq \sum_{i}^{n}x_{i}^{2}/G$

By using Cauchy-Schwarz' Inequality.

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  • $\begingroup$ The book has not yet introduced that inequality. Is there any other way to prove it? The text before this exercise is about properties of the integral and average value. I was thinking about this: "If $l(x) \le v(x) \le u(x)$ for $a \le x \le b$, then $\int_a^bl(x)dx \le \int_a^bv(x)dx \le \int_a^bu(x)dx$", but it does not seem to fit in proof. $\endgroup$
    – W. Zhu
    Commented Dec 13, 2016 at 13:35

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