Inequality regarding sample mean

Question

I was looking at the book "Asymptotic Theory of statistics and probability, DasGupta A., 2008" and in one point of a proof they use an inequality which I have not been able to understand.

Given that $X_i, i \in \{1,...,n\}$ are independent and identically distributed random variables, with mean $\mu$, and with sample mean $\overline{X}_n $, they state that

$$ \sum_{i=1}^n |X_i - \overline{X}_n|^3 \leq 2³\left( \sum_{i=1}^n |X_i - \mu|^3 + n|\mu - \overline{X}_n|^3 \right)$$

I don't know how the 2³ term appears. I've tried adding and substracting $\mu$ but I have not been able to proof the inequality. It would be perfect if you could lend me a hand.

Thanks

Answer 1

For $a$ and $b$ non-negative you have $$(a+b)^3 \le (2\max(a,b))^3 = 2^3\max(a,b)^3 \le 2^3 (a^3+b^3)$$

So here you can say $$\sum\limits_{i=1}^n \bigg|X_i - \overline{X}_n\bigg|^3 \\ = \sum\limits_{i=1}^n \bigg|(X_i -\mu) + (\mu-\overline{X}_n)\bigg|^3 \\ \le \sum\limits_{i=1}^n \bigg( |X_i -\mu| + |\mu-\overline{X}_n|\bigg)^3 \\ \le \sum\limits_{i=1}^n 2^3\bigg( |X_i -\mu|^3 + |\mu-\overline{X}_n|^3\bigg) \\ = 2^3\bigg(\bigg( \sum\limits_{i=1}^n |X_i -\mu|^3\bigg) + \bigg( \sum\limits_{i=1}^n|\mu-\overline{X}_n|^3\bigg)\bigg)\\ = 2^3\bigg(\bigg( \sum\limits_{i=1}^n |X_i -\mu|^3\bigg) + n|\mu-\overline{X}_n|^3\bigg)$$

Answer 2

The inequality is ugly but it is true. First call $X_i-\mu=a_i$ and $\mu-\bar X_n=b$. If $b=0$ it is trivial. Otherwise divide by $|b|^3$ both sides. Your inequality now reads $$ \sum|a_i/b+1|^3\le 8(\sum|a_i/b|^3+n) $$ It is enough to prove that $$ |t+1|^3\le 8(|t|^3+1) $$ For all $t$ (You can then plug in $t=a_i/b$ and add them up). Consider the cases one by one to remove the absolute value. If $t>0$ your inequality reduces to $$ 7t^2+6t+7\ge 0 $$ Which is clearly true for all $t$. If $-1\le t\le 0$ it reduces to $$ 9t^3+3t^2+3t-7\le 0 $$ Which is true even without the $-7$, and if $t\le -1$ it reduces to $$ 7t^2+10t+7\ge 0, $$ Which is true for all $t$ as well.