[proofplan] We first use the elementary pointwise bound \begin{align*} |x| \leq \frac{x^2+1}{2} \end{align*} to prove that $|X|$ has finite expectation. Once $\mathbb E[|X|]$ is known to be finite, we can safely expand the non-negative square $(|X|-t)^2$ and use it to obtain a quadratic inequality in $t$. Choosing the optimal value $t=\mathbb E[|X|]$ gives $\mathbb E[|X|]^2 \leq \mathbb E[X^2]$, which is the desired estimate. [/proofplan] [step:Use the quadratic pointwise bound to prove integrability] For every $x \in \mathbb R$, \begin{align*} 0 \leq (|x|-1)^2 = x^2 - 2|x| + 1, \end{align*} and hence \begin{align*} |x| \leq \frac{x^2+1}{2}. \end{align*} Applying this pointwise inequality to $x=X(\omega)$ for $\omega \in \Omega$ gives \begin{align*} |X(\omega)| \leq \frac{X(\omega)^2+1}{2}. \end{align*} Since $X^2$ is integrable by hypothesis and $\mathbb P(\Omega)=1$, we obtain \begin{align*} \mathbb E[|X|] = \int_\Omega |X(\omega)|\,d\mathbb P(\omega) \leq \frac{1}{2}\int_\Omega X(\omega)^2\,d\mathbb P(\omega) + \frac{1}{2}\int_\Omega 1\,d\mathbb P(\omega) = \frac{\mathbb E[X^2]+1}{2} < \infty. \end{align*} Thus $X \in L^1(\Omega,\mathcal F,\mathbb P)$. [guided] The first issue is not the sharp inequality, but the basic fact that $\mathbb E[|X|]$ is finite. We cannot manipulate $\mathbb E[|X|]$ as a real number until we have proved this. The elementary estimate we need comes from a non-negative square: for every $x \in \mathbb R$, \begin{align*} 0 \leq (|x|-1)^2 = x^2 - 2|x| + 1. \end{align*} Rearranging this inequality gives \begin{align*} |x| \leq \frac{x^2+1}{2}. \end{align*} Now substitute $x=X(\omega)$ for each $\omega \in \Omega$. Since $X$ is a real-valued [random variable](/page/Random%20Variable), this gives the pointwise estimate \begin{align*} |X(\omega)| \leq \frac{X(\omega)^2+1}{2}. \end{align*} The right-hand side is integrable with respect to $\mathbb P$: the term $X^2$ is integrable by the square-integrability hypothesis, and the constant function $1: \Omega \to \mathbb R$ is integrable because $\mathbb P$ is a probability measure and $\mathbb P(\Omega)=1$. Therefore \begin{align*} \mathbb E[|X|] = \int_\Omega |X(\omega)|\,d\mathbb P(\omega) \leq \frac{1}{2}\int_\Omega X(\omega)^2\,d\mathbb P(\omega) + \frac{1}{2}\int_\Omega 1\,d\mathbb P(\omega) = \frac{\mathbb E[X^2]+1}{2} < \infty. \end{align*} This proves that $X$ is integrable, so $\mathbb E[|X|]$ is now a finite real number. [/guided] [/step] [step:Expand a non-negative square to obtain the sharp estimate] Define the finite numbers \begin{align*} A := \mathbb E[|X|], \qquad B := \mathbb E[X^2]. \end{align*} For each $t \in \mathbb R$, the random variable \begin{align*} \Omega &\to \mathbb R \\ \omega &\mapsto (|X(\omega)|-t)^2 \end{align*} is non-negative. It is also integrable, because the pointwise identity \begin{align*} (|X(\omega)|-t)^2 = X(\omega)^2 - 2t|X(\omega)| + t^2 \end{align*} expresses it as a linear combination of the integrable random variables $X^2$, $|X|$, and the constant function $1: \Omega \to \mathbb R$. Therefore its expectation is non-negative. Expanding the square and using the finiteness of $A$ and $B$ gives \begin{align*} 0 \leq \int_\Omega (|X(\omega)|-t)^2\,d\mathbb P(\omega) = B - 2tA + t^2. \end{align*} Taking $t=A$ yields \begin{align*} 0 \leq B - A^2. \end{align*} Thus \begin{align*} \mathbb E[|X|]^2 \leq \mathbb E[X^2]. \end{align*} Since $\mathbb E[|X|]\geq 0$ and $\mathbb E[X^2]\geq 0$, taking square roots gives \begin{align*} \mathbb E[|X|] \leq \sqrt{\mathbb E[X^2]}. \end{align*} This proves the claimed inequality. [/step]