# 113 新北新莊高中 校內資訊學科能力競賽 p8

#### Proposition:

Let $a_1, a_2, \dots, a_n$ be $n$ real numbers. Define the variance as $$V=\frac{1}{n}\sum_{i=1}^{n}{(a_i-\mu)^2}$$, where $\mu$ is the arithmetic mean, $$\mu=\frac{1}{n}\sum_{i=1}^{n}{a_i}$$.

Similarly, define the **super variance** as $$sV=\frac{1}{n}\sum_{i=1}^{n}{(a_i-g)^2}$$, where $g$ is the geometric mean, $$g=(\prod_{i=1}^{n}{a_i})^{\frac{1}{n}}$$.

We aim to compare $V$ and $sV$.

#### Proof:
Expanding the expressions for $V$ and $sV$, we obtain:
$$V=\frac{1}{n}\sum_{i=1}^{n}{(a_i-\mu)^2}=\frac{1}{n}\sum_{i=1}^{n}{({a_i}^2-2a_i\mu+\mu^2)}$$. Since $\sum_{i=1}^{n}{a_i}=n\mu$, we simplify:
$$V=\frac{1}{n}(\sum_{i=1}^{n}{{a_i}^2}-2(n\mu)\mu+n\mu^2)=\frac{1}{n}\sum_{i=1}^{n}{{a_i}^2}-\mu^2$$.

Similarly, for $sV$:
\begin{aligned}
sV&=\frac{1}{n}\sum_{i=1}^{n}{(a_i-g)^2}\\
&=\frac{1}{n}\sum_{i=1}^{n}{({a_i}^2-2a_ig+g^2)}\\
&=\frac{1}{n}(\sum_{i=1}^{n}{{a_i}^2}-2(n\mu)g+ng^2)\\
&=\frac{1}{n}\sum_{i=1}^{n}{{a_i}^2}-(2\mu g-g^2)
\end{aligned}.

To compare $V$ and $sV$, we analyze the term:
\begin{aligned}
\mu^2-(2\mu g-g^2)&=\mu^2-2\mu g+g^2\\
&=(\mu-g)^2
\end{aligned}.
By the **AM-GM inequality**, we know $\mu\ge g$, with equality holding if and only if $a_1=a_2=\cdots=a_n$. Since $(\mu-g)^2\ge0$, it follows that $$\mu^2\ge2\mu g-g^2$$.

Thus, we conclude $$V=\frac{1}{n}\sum_{i=1}^{n}{{a_i}^2}-\mu^2\le\frac{1}{n}\sum_{i=1}^{n}{{a_i}^2}-(2\mu g-g^2)=sV$$. Equality holds if and only if $a_1=a_2=\cdots=a_n$.

Therefore,
$$V\le sV,\ \text{with equality holding if and only if }a_1=a_2=\cdots=a_n\text .$$
$\blacksquare$