[proofplan]
We center the two random variables and reduce the covariance estimate to a purely quadratic inequality for two square-integrable real-valued random variables. The key estimate is obtained by applying non-negativity of $\mathbb E[(U+\lambda V)^2]$ for every real parameter $\lambda$, which forces the discriminant of the associated quadratic polynomial to be non-positive. Once $|\operatorname{Cov}(X,Y)|\le \sigma_X\sigma_Y$ is established, the bound on the correlation coefficient follows by dividing by the positive product $\sigma_X\sigma_Y$.
[/proofplan]
[step:Center the variables and record square integrability]
Let $m_X:=\mathbb E[X]$ and $m_Y:=\mathbb E[Y]$. These expectations are finite because, for every $t\in\mathbb R$,
\begin{align*}
|t|\le \frac{1+t^2}{2},
\end{align*}
and hence
\begin{align*}
\mathbb E[|X|]\le \frac{1+\mathbb E[X^2]}{2}<\infty,\qquad
\mathbb E[|Y|]\le \frac{1+\mathbb E[Y^2]}{2}<\infty.
\end{align*}
Define the centered random variables
\begin{align*}
U:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)),&
\omega&\mapsto X(\omega)-m_X,\\
V:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)),&
\omega&\mapsto Y(\omega)-m_Y.
\end{align*}
Since $(a-b)^2\le 2a^2+2b^2$ for all $a,b\in\mathbb R$, we have
\begin{align*}
\mathbb E[U^2]\le 2\mathbb E[X^2]+2m_X^2<\infty,\qquad
\mathbb E[V^2]\le 2\mathbb E[Y^2]+2m_Y^2<\infty.
\end{align*}
Thus
\begin{align*}
\operatorname{Cov}(X,Y)=\mathbb E[UV],\qquad
\sigma_X^2=\mathbb E[U^2],\qquad
\sigma_Y^2=\mathbb E[V^2].
\end{align*}
The product $UV$ is integrable because
\begin{align*}
2|UV|\le U^2+V^2,
\end{align*}
so $\mathbb E[UV]$ is well-defined.
[/step]
[step:Force the quadratic discriminant to be non-positive]
Set
\begin{align*}
a:=\mathbb E[U^2],\qquad b:=\mathbb E[UV],\qquad c:=\mathbb E[V^2].
\end{align*}
For each $\lambda\in\mathbb R$, define
\begin{align*}
Z_\lambda:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)),&
\omega&\mapsto U(\omega)+\lambda V(\omega).
\end{align*}
The [random variable](/page/Random%20Variable) $Z_\lambda$ is square-integrable because, for every $\omega\in\Omega$,
\begin{align*}
Z_\lambda(\omega)^2\le 2U(\omega)^2+2\lambda^2V(\omega)^2,
\end{align*}
and the right-hand side has finite expectation. Hence $Z_\lambda^2$ is integrable. Since $Z_\lambda^2\ge 0$,
\begin{align*}
0\le \mathbb E[Z_\lambda^2]
= \mathbb E[U^2]+2\lambda\mathbb E[UV]+\lambda^2\mathbb E[V^2]
= a+2\lambda b+\lambda^2 c.
\end{align*}
If $c=0$, then $\mathbb E[V^2]=0$. Since $V^2\ge 0$, this implies $V=0$ $\mathbb P$-a.s.; hence $UV=0$ $\mathbb P$-a.s. and $b=0$. Therefore
\begin{align*}
b^2=0=ac.
\end{align*}
If $c>0$, choose $\lambda=-b/c$. Substituting this value into the non-negative quadratic gives
\begin{align*}
0\le a+2\left(-\frac{b}{c}\right)b+\left(-\frac{b}{c}\right)^2c
= a-\frac{b^2}{c}.
\end{align*}
Multiplying by $c>0$ yields
\begin{align*}
b^2\le ac.
\end{align*}
Thus, in all cases,
\begin{align*}
|\mathbb E[UV]|\le \sqrt{\mathbb E[U^2]}\sqrt{\mathbb E[V^2]}.
\end{align*}
[guided]
The goal is to prove a Cauchy-Schwarz-type estimate without assuming it as a separate theorem. The standard trick is to use the fact that the square of any real-valued random variable has non-negative expectation.
Define
\begin{align*}
a:=\mathbb E[U^2],\qquad b:=\mathbb E[UV],\qquad c:=\mathbb E[V^2].
\end{align*}
These quantities are finite: $a$ and $c$ are finite by square integrability of $U$ and $V$, while $b$ is finite because
\begin{align*}
2|UV|\le U^2+V^2.
\end{align*}
For a real parameter $\lambda\in\mathbb R$, define the random variable
\begin{align*}
Z_\lambda:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)),&
\omega&\mapsto U(\omega)+\lambda V(\omega).
\end{align*}
Before expanding $\mathbb E[Z_\lambda^2]$, we verify that this expectation is finite. For every $\omega\in\Omega$,
\begin{align*}
Z_\lambda(\omega)^2\le 2U(\omega)^2+2\lambda^2V(\omega)^2,
\end{align*}
so square integrability of $U$ and $V$ implies $\mathbb E[Z_\lambda^2]<\infty$. Since $Z_\lambda^2\ge 0$ pointwise, its expectation is non-negative:
\begin{align*}
0\le \mathbb E[Z_\lambda^2].
\end{align*}
Expanding the square and using linearity of expectation gives
\begin{align*}
0\le \mathbb E[(U+\lambda V)^2]
= \mathbb E[U^2]+2\lambda\mathbb E[UV]+\lambda^2\mathbb E[V^2]
= a+2\lambda b+\lambda^2 c.
\end{align*}
This inequality holds for every $\lambda\in\mathbb R$. If $c=0$, then $\mathbb E[V^2]=0$. Because $V^2$ is non-negative, this forces $V=0$ $\mathbb P$-a.s.; consequently $UV=0$ $\mathbb P$-a.s. and $b=\mathbb E[UV]=0$. Hence
\begin{align*}
b^2=0=ac.
\end{align*}
Now suppose $c>0$. Since the quadratic expression
\begin{align*}
a+2\lambda b+\lambda^2 c
\end{align*}
is non-negative for every real $\lambda$, evaluate it at the minimizing value $\lambda=-b/c$. This gives
\begin{align*}
0\le a+2\left(-\frac{b}{c}\right)b+\left(-\frac{b}{c}\right)^2c
= a-\frac{b^2}{c}.
\end{align*}
Multiplying by the positive number $c$ preserves the inequality and yields
\begin{align*}
b^2\le ac.
\end{align*}
Substituting back $a=\mathbb E[U^2]$, $b=\mathbb E[UV]$, and $c=\mathbb E[V^2]$, we obtain
\begin{align*}
|\mathbb E[UV]|\le \sqrt{\mathbb E[U^2]}\sqrt{\mathbb E[V^2]}.
\end{align*}
[/guided]
[/step]
[step:Translate the centered estimate into the covariance bound]
Using $U=X-\mathbb E[X]$ and $V=Y-\mathbb E[Y]$, the estimate from the previous step gives
\begin{align*}
|\operatorname{Cov}(X,Y)|
=|\mathbb E[UV]|
\le \sqrt{\mathbb E[U^2]}\sqrt{\mathbb E[V^2]}
=\sqrt{\operatorname{Var}(X)}\sqrt{\operatorname{Var}(Y)}
=\sigma_X\sigma_Y.
\end{align*}
This proves the covariance bound.
[/step]
[step:Divide by the product of standard deviations]
Assume $\operatorname{Var}(X)>0$ and $\operatorname{Var}(Y)>0$. Then $\sigma_X>0$ and $\sigma_Y>0$, so $\sigma_X\sigma_Y>0$. Dividing the covariance bound by this positive number gives
\begin{align*}
\left|\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}\right|\le 1.
\end{align*}
By the definition of $\rho(X,Y)$, this is
\begin{align*}
|\rho(X,Y)|\le 1,
\end{align*}
which is equivalent to
\begin{align*}
-1\le \rho(X,Y)\le 1.
\end{align*}
This completes the proof.
[/step]
