How can I prove $(2\sum_{i}^{n}{x_i})^2 - 4(n\sum_{i=1}^{n}{x_i^2})$ is negative?

Question

I am having some difficulty proving that $(2\sum_{i}^{n}{x_i})^2 - 4(n\sum_{i=1}^{n}{x_i^2})$ is a negative number, and thus that the residual sum of squares (RSS) is an ellipse.

I used this link to show that RSS is a conic section and was successfully able to do so. I just am struggling to prove that the above subtraction is negative.

Whether or not this is related, this form looks vastly similar to \begin{align} \mathrm{Var}(x) &= \mathbb{E}(x^2) - \mathbb{E}(x)^2 \geq 0 \\ &=\frac{\sum_{i=1}^{n}{x_i^2}}{n} - \frac{(\sum_{i=1}^{n}{x_i})^2}{n^2} \geq 0 \end{align}

Then $\frac{\sum_{i=1}^{n}{x_i^2}}{n} \geq \frac{(\sum_{i=1}^{n}{x_i})^2}{n^2}$.

Could I use this to rewrite $(2\sum_{i}^{n}{x_i})^2 - 4(n\sum_{i=1}^{n}{x_i^2})$ as \begin{align} &= 4n^2[\frac{(\sum_{i=1}^{n}{x_i})^2}{n^2}-\frac{(\sum_{i=1}^{n}{x_i^2})}{n}] \end{align} or is this wrong?

Answer 1

Your approach is right, but for the wrong reasons (as you've written it) : the law of large numbers tells you that the sample mean $\bar x$ converges to the expectation of $x$ as the sample size $n$ goes to infinity, but the sample mean is not equal to the true expectation : $$\mathbb E(x)\mathbf\ne\ \bar x =\frac{\sum_{i=1}^{n}{x_i}}{n}\quad \text{ and } \quad\mathbb E(x^2)\mathbf\ne\frac{\sum_{i=1}^{n}{x_i^2}}{n} $$

(Assuming that all the $x_i$ are i.i.d. following the same distribution as $x$)

However, as maybe you wanted to write, the quantity $\frac{\sum_{i=1}^{n}{x_i^2}}{n} - \frac{(\sum_{i=1}^{n}{x_i})^2}{n^2}$ is equal to the sample variance

$$\bar s := \frac{1}{n} \sum_{i=1}^n (x_i-\bar{x})^2 \ge 0 $$

and is therefore non-negative (if you're not sure why, just develop $\bar s$). So using that argument, your desired inequality holds.