Proof that if higher moment exists then lower moment also exists The $r$-th moment of a random variable $X$ is finite if
$$
\mathbb E(|X^r|)< \infty
$$
I am trying to show that for any positive integer $s<r$, then the
$s$-th moment $\mathbb E[|X^s|]$ is also finite.
 A: $0<s<r \Longrightarrow \forall X \, |X|^s \le \max(1, |X|^r) $
A: You can prove it with the help of the monotonicity property of integration,i.e., for functions $f$ and $g$ such that $f \geqslant g$, then $\mathbb{E}[f(X)] \geqslant \mathbb{E}[g(X)]$.
Proof:
Consider the function $g:\mathbb{R} \to \mathbb{R}$ such that $g(x) = |x|^r + 1$ and the function $h: \mathbb{R} \to \mathbb{R}$ such that $h(x) = |x|^s$. Also, consider that $F$ is the cdf of $X$. We can notice that, for every $r \geqslant s$, $g(x) > h(x)$. Then, by monotonicity of integrals, we have
$$
    \int_{\mathbb{R}} |x|^r + 1 \, dF = \int_{\mathbb{R}} |x|^r\, dF + \underbrace{\int_{\mathbb{R}} 1 \, dF}_{=1} > \int_{\mathbb{R}} |x|^s \, dF
    $$
Since $\int_{\mathbb{R}} |x|^r\, dF = \mathbb{E}|X|^{r}<\infty$, then
$$
    \infty > \mathbb{E}|X|^{r} + 1 > \int_{\mathbb{R}} |x|^s \, dF = \mathbb{E}|X|^{s}
$$

There are some comments about using Jensen's Inequality to prove it, but it is wrong to use it, since it becomes a circular proof. From Casella & Berger:
Theorem 4.7.7 (Jensen's Inequality) For any random variable $X$, if $g(x)$ is a convex function, then
$$
\mathbb{E}[g(X)] \geqslant g(\mathbb{E}[X])
$$
This means that, to use this inequality, we must know that $\mathbb{E}[X] < \infty$.
In our case, by applying a function $\varphi: \mathbb{R} \to \mathbb{R}$ such that $\varphi(x) = x^{\frac{r}{s}}$ on $\mathbb{E}[|X|^s]$, we are assuming that $\mathbb{E}[|X|^s] < \infty$, which is exactly what we want to prove.
