prove $\sum_{i=1}^{n}(x_i-\bar{x})^2\lt\sum_{i=1}^{n}(x_i-a)^2$

Question

How do we prove that $$ \sum_{i=1}^{n}(x_i-\bar{x})^2\lt\sum_{i=1}^{n}(x_i-a)^2 $$

where $a$ is any value other than $\bar{x}$, the arithmetic mean.

My Attempt: $$ \sum_{i=1}^{n}(x_i^2-2x_i\bar{x}+\bar{x}^2)\lt\sum_{i=1}^{n}(x_i^2-2x_ia+a^2)\\\sum_{i=1}^{n}x_i^2-2\bar{x}\sum_{i=1}^{n}x_i+\sum_{i=1}^{n}\bar{x}^2\lt\sum_{i=1}^{n}x_i^2-2a\sum_{i=1}^{n}x_i+\sum_{i=1}^{n}a^2\\-2\bar{x}.n\bar{x}+n.\bar{x}^2\lt-2a.n\bar{x}+n.a^2\\-2\bar{x}^2+\bar{x}^2\lt-2a\bar{x}+a^2\\-\bar{x}^2\lt-2a\bar{x}+a^2 $$ But how do I proceed further or is there any better way ?

Answer 1

consider $$f(a) \equiv\sum_{i=1}^{n}(x_i-a)^2 $$

so that $$f'(a) =-2\sum_{i=1}^{n}(x_i-a) =-2n(\bar x -a) $$ so $f(a)$ is minimized when $a=\bar x$

Answer 2

Once you have gotten the last intequality, just note that $$-\bar{x}^2<-2a\bar{x}+a^2\iff \bar{x}^2 -2a\bar{x}+a^2\iff(\bar{x}-a)^2>0.$$

Answer 3

If $a \neq \bar{x}$,

$$(a-\bar{x})^2 >0$$

$$a^2+\bar{x}^2-2a\bar{x}>0$$

$$-\bar{x}^2<a^2-2a\bar{x}$$

Answer 4

It would be simpler to think this as a quadratic function of $a$: $$ f(a) = \sum_{i=1}^n (x_i-a)^2 = A a^2 + B a + C, $$ where $$ A = n, \quad B = -2 \sum_{i=1}^n x_i, \quad\text{and}\quad C = \sum_{i=1}^n x_i^2. $$ "Completing the square" shows that for $A > 0$ the parabola $f(a)$ is minimized at $a = -B/(2A)$. Substituting the values above yields a minimum at $a = \bar{x}$.

Answer 5

$\begin{array}\\ \sum_{i=1}^{n}(x_i-a)^2-\sum_{i=1}^{n}(x_i-\bar{x})^2 &=\sum_{i=1}^{n}((x_i-a)^2-(x_i-\bar{x})^2)\\ &=\sum_{i=1}^{n}((x_i-a)-(x_i-\bar{x}))((x_i-a)+(x_i-\bar{x}))\\ &=\sum_{i=1}^{n}(\bar{x}-a)(2x_i-(a+\bar{x}))\\ &=\sum_{i=1}^{n}(\bar{x}-a)(2x_i)-\sum_{i=1}^{n}(\bar{x}-a)(a+\bar{x})\\ &=2(\bar{x}-a)\sum_{i=1}^{n}x_i-n(\bar{x}^2-a^2)\\ &=2(\bar{x}-a)n\bar{x}-n(\bar{x}^2-a^2)\\ &=2n\bar{x}^2-2an\bar{x}-n\bar{x}^2+na^2\\ &=n\bar{x}^2-2an\bar{x}+na^2\\ &=n(\bar{x}^2-2a\bar{x}+a^2)\\ &=n(\bar{x}-a)^2\\ &\ge 0\\ \end{array} $

with equality only if $a = \bar{x}$.

Answer 6

By factoring the difference of squares and taking the means (proportional to the sums),

$$\overline{(x_i-\overline x)^2-(x_i-a)^2}=\overline{(x_i-\overline x+x_i-a)}\overline{(a-\overline x)}=(\overline x-\overline x+\overline x-a)(a-\overline x)<0.$$