Proof of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm

Throughout we use the linearity and monotonicity of expectation from Linearity and Monotonicity of the Lebesgue Integral and the closure properties of square-integrable random variables from that definition (sums and scalar multiples are square-integrable; products are integrable; in particular all expectations below are defined and finite).

Part 1. Write $a=\lVert X\rVert_{2}^{2}=\mathbb{E}[X^{2}]$, $b=\mathbb{E}[XY]$, and $c=\lVert Y\rVert_{2}^{2}=\mathbb{E}[Y^{2}]$.

Case $c>0$. For every real $t$, the random variable $(X+tY)^{2}$ is nonnegative, and expanding pointwise, $(X+tY)^{2}=X^{2}+2tXY+t^{2}Y^{2}$, so by linearity and monotonicity of expectation

Taking $t=-b/c$ gives $0\le a-2b^{2}/c+b^{2}/c=a-b^{2}/c$, hence $b^{2}\le ac$. Since the nonnegative square root is order preserving on nonnegative reals (if $0\le u\le v$ and $\sqrt{u}>\sqrt{v}$ then squaring the inequality of nonnegative numbers would give $u>v$), we get $|b|=\sqrt{b^{2}}\le\sqrt{a}\sqrt{c}=\lVert X\rVert_{2}\lVert Y\rVert_{2}$, using $\sqrt{ac}=\sqrt{a}\sqrt{c}$ (both sides are nonnegative with square $ac$, so they agree by the uniqueness in the cited theorem) and $\sqrt{b^{2}}=|b|$ (both sides are nonnegative with square $b^{2}$).

Case $c=0$. By Square-Integrable Random Variables and the Mean-Square Inner Product, $\lVert Y-0\rVert_{2}=0$ implies $P(Y=0)=1$, hence $P(XY=0)=1$ since $\{Y=0\}\subseteq\{XY=0\}$ and probabilities are monotone. Then $\mathbb{E}[|XY|]=0$: every nonnegative simple function $s\le|XY|$ vanishes on the event $\{XY=0\}$, so each of its finitely many nonzero values is attained on a subset of the event $\{XY\ne0\}$ of probability $0$, giving $s$ integral $0$ by monotonicity of the measure $P$; the integral of the nonnegative function $|XY|$ is the supremum of these, hence $0$. By Linearity and Monotonicity of the Lebesgue Integral, $|\mathbb{E}[XY]|\le\mathbb{E}[|XY|]=0$, so both sides of the claimed inequality are $0$ and it holds.

Part 2. Expanding pointwise and using linearity,

where the middle step is Part 1 together with $\mathbb{E}[XY]\le|\mathbb{E}[XY]|$. Both $\lVert X+Y\rVert_{2}$ and $\lVert X\rVert_{2}+\lVert Y\rVert_{2}$ are nonnegative, so the order-preservation of the nonnegative square root recalled in Part 1 yields $\lVert X+Y\rVert_{2}\le\lVert X\rVert_{2}+\lVert Y\rVert_{2}$. $\blacksquare$