Preface: please talk to me like a human being and not in esoteric mathematical language you assume I understand :) My highest level of math was calculus 3 / linear algebra.
Okay, my question is very simple: I am trying to prove Chebyshev's inequality by following along in the wiki article. And I understood most of it, except the following step:
Basically, I want to understand why the inequality sign flipped in one case but didn't in the other.
On
Squaring stuff preserves order from $0$, in the following sense: If $a$ is further from $0$ than $b$ is from $0$, $a^2$ is further from $0$ than $b^2$ is from $0$.
When $x\in R_1$, we have $x-\mu<-k\sigma\le0$ (I assume $k$ is a nonnegative integer). Here, $k\sigma$ is closer to $0$ than $x-\mu$, so $k^2\sigma^2$ is closer to $0$ than $(x-\mu)^2$ is to $0$. Both numbers are nonnegative since they're squares, so we have $0\le k^2\sigma^2<(x-\mu)^2$.
When $x\in R_3$, we have $x-\mu>k\sigma\ge0$. Similar reasoning tells us $k^2\sigma^2$ is closer to $0$ than $(x-\mu)^2$ is to $0$, but the numbers were already positive so squaring doesn't switch signs.
Using the identity $\,a^2-b^2=(a-b)(a+b)\,$ you get that: $$(x-\mu)^2-k^2 \sigma^2 = (x-\mu - k\sigma)(x-\mu + k\sigma) \tag{1}$$
The context assumes that $\,k, \sigma \ge 0\,$. Then:
1st case $\,x \lt \mu - k \sigma \iff x - \mu + k \sigma \lt 0$ implies that $x -\mu-k\sigma \le x - \mu + k \sigma \lt 0\,$, therefore the RHS of $\,(1)\,$ is a product of two negative numbers, so the LHS is positive $\,(x-\mu)^2-k^2 \sigma^2 \gt 0 \iff (x-\mu)^2 \gt k^2 \sigma^2\,$;
2nd case $\,x \gt \mu + k \sigma \iff x - \mu - k \sigma \gt 0$ implies that $x -\mu + k\sigma \ge x - \mu - k \sigma \gt 0\,$, therefore the RHS of $\,(1)\,$ is a product of two positive numbers, so the LHS is again positive.