In the Appendix A of this book: Statistics: Principles and Methods written by Giuseppe Cicchitelli, Pierpaolo D’Urso and Marco Minozzo published by Pearson in 2021, I found the most simple proof of the Chebyshev theorem I have ever seen. The idea of using two different sets of element is very smart.

I reproduced the demonstration here:

Let *x _{1}*,

*x*, …,

_{2}*x*be a series of observations with mean

_{N}*µ*and variance

*σ*. Let

^{2}I_k=[i, 1 \leq i \leq N:\left|x_i-\mu\right|< k\sigma ]

be the set of subscripts *i* identifying the observations whose deviation from the mean is (in absolute value) less than *kσ*. Let *N(I _{k})* be the number of elements in

*I*. We can write

_{k}\begin{aligned} \sigma^2 &=\dfrac{\sum_{i=1}^N\left(x_i-\mu\right)^2}{N } \\ N \sigma^2 &=\sum_{i=1}^N\left(x_i-\mu\right)^2 \\ &=\sum_{i \in I_k}\left(x_i-\mu\right)^2+\sum_{i \notin I_k}\left(x_i-\mu\right)^2 \\ & \geq \sum_{i \notin I_k}\left(x_i-\mu\right)^2 \\ & \geq \sum_{i \notin I_k} k^2 \sigma^2 \end{aligned}

where the first inequality holds because the sum of squared deviations from the mean extends over the subset of *x _{i}* not belonging to

*I*, while the second holds since

_{k}*(xi-µ)*.

^{2}>k^{2}σ^{2}This last point can be illustrated by numerical values. Indeed, with µ=0, *σ*=1, and *k*=2, we have for *(xi-µ) ≤k σ* :

For *(xi-µ) ^{2} >k^{2} σ^{2}*, we have:

Hence,

\sum_{i \notin I_k} k^2 \sigma^2 \leq N \sigma^2,

from which, by dividing both sides of the inequality by *N k ^{2} σ^{2}*, we obtain

\frac{1}{N} \sum_{i \notin I_k}(1) \leq \frac{1}{k^2} \Leftrightarrow \frac{N-N\left(I_k\right)}{N} \leq \frac{1}{k^2} .

Recall that the total number of element *N* is equal to the* *number of elements in *Ik,* *N(Ik)* plus the number of elements that are not in *Ik*:

N=N\left(I_k\right)+N\left(\text{not }I_k\right)\\ \sum_{i \notin I_k}(1)=N\left(\text{not }I_k\right)=N-N\left(I_k\right)

Finally,

\frac{N\left(I_k\right)}{N} \geq 1-\frac{1}{k^2}

and the result is proved.