Prove: $E[|XY|] \le \sqrt{E[X^2] + E[Y^2]}$
Starting with:
$$E\Bigg[\Big(\alpha~|X| - |Y|\Big)^2\Bigg] \ge 0$$
$$E\Big[\alpha^2~|X|^2\Big] - 2~\alpha~ E\Big[~|XY|~\Big] + E\Big[Y^2\Big] \ge 0$$
$$\alpha^2 ~E\Big[X^2\Big] - 2~ \alpha~ E\Big[~|XY|~\Big] + E\Big[Y^2\Big] \ge 0$$
I'm not sure what they did to get to the next step:
$$E\Bigg[~2~ \Big|XY\Big|~ \Bigg]^2 -4~ E[X^2]~ E[Y^2] \le 0$$
Any Ideas?
Their explanation is something like this: "The discriminant of the quadratic in $\alpha$ appearing in equation must be non-positive, because the quadratic cannot have two distict roots. Therefore:
$$E\Bigg[~2~ \Big|XY\Big|~ \Bigg]^2 -4~ E[X^2]~ E[Y^2] \le 0$$
$$ E[|XY|]^2 \le E[X^2]~E[Y^2]$$
$$E[|XY|] \le \sqrt{E[X^2]~E[Y^2]}$$
$$ f(\alpha)=\alpha^2 ~E\Big[X^2\Big] - 2~ \alpha~ E\Big[~|XY|~\Big] + E\Big[Y^2\Big] \ge 0 \tag{1} $$ here you have quadratic function, so it's discriminant is $$ \Delta=b^2-4ac=E\Bigg[~2~ \Big|XY\Big|~ \Bigg]^2 -4~ E[X^2]~ E[Y^2]. $$ By (1), $f$ has at most one root, so $\Delta\leq0$.