Note that we alternatively could prove this inequality by showing that the \(\langle X, Y \rangle = \mathbb{E}[XY]\) is an inner product space and then cite the C-S inequality.
Suppose we have random variables \(X\) and \(Y\) where \(\mathbb{E}[X^2]\) and \(\mathbb{E}[Y^2]\) are finite. Pick any \(t \in \mathbb{R}\). It is obvious that \(\mathbb{E}[(tX + Y)^2] \geq 0\) and so we have:
\[\mathbb{E}[(tX + Y)^2] = \mathbb{E}[X^2]t^2 + 2\mathbb{E}[XY]t + \mathbb{E}[Y^2] \geq 0\]which is a quadratic in \(t\). Note that because we are sure for any \(t \in \mathbb{R}, \mathbb{E}[(tX + Y)^2] \geq 0 \implies\) the discriminant (i.e. \(b^2 - 4ac\)) of the above quadratic is \(\leq 0\). Using this condition for our quadratic above we have:
\[4\mathbb{E}[XY]^2 - 4\mathbb{E}[X^2]\mathbb{E}[Y^2] \leq 0 \implies \boxed{\mathbb{E}[XY]^2 \leq \mathbb{E}[X^2]\mathbb{E}[Y^2]}\]