$\text{Cov}(\max(X,Y), \max(X, Z)) \leq \text{Var}(X)$ Given three independent random variables $X, Y, Z$, is it true that
$$\text{Cov}(\max(X,Y), \max(X, Z)) \leq C \cdot \text{Var}(X)?$$
I am thinking about the case where the $\text{Var}(Y)$ and $\text{Var}(Z)$ are much larger than $\text{Var}(X)$, so Cauchy-Schwartz inequality would not be useful here.
My gut feeling this should hold because of the independence. I tried a few conditioning. For example, we know $\max(X, Z)\perp Y$, so if we condition on $Y$, we have
$$\text{Cov}(\max(X,Y), \max(X, Z)) = \mathbb{E}[\text{Cov}_{\mathcal{F}_Y}(\max(X,Y), \max(X, Z))].$$
I also tried conditioning on $X$, but these did not lead to anything at this moment.
 A: Write $\widetilde{Y} = \max\{Y, X\}$ and $\widetilde{Z} = \max\{Z, X\}$ for brevity. Then by the law of total covariance,
\begin{align*}
\mathbf{Cov}(\widetilde{Y}, \widetilde{Z})
&= \mathbf{E}[\overbrace{ \mathbf{Cov}(\widetilde{Y}, \widetilde{Z} \mid X) }^{=0}] + \mathbf{Cov}(\mathbf{E}[\widetilde{Y} \mid X], \mathbf{E}[\widetilde{Z} \mid X]) \\
&= \mathbf{Cov}(f(X), g(X)),
\end{align*}
where $f(\cdot)$ and $g(\cdot)$ are defined by
\begin{align*}
f(x) &= \mathbf{E}[\widetilde{Y} \mid X = x] = \mathbf{E}[\max\{Y, x\}], \\
g(x) &= \mathbf{E}[\widetilde{Z} \mid X = x] = \mathbf{E}[\max\{Z, x\}].
\end{align*}
Now we note that both $f$ and $g$ are $1$-Lipschitz, since $x \mapsto \max\{y, x\}$ is $1$-Lipschitz for any $y \in \mathbb{R}$. To make use of this observation, we need the following result:

Lemma. Let $X \in L^2(\mathbf{P})$, and let $f : \mathbb{R} \to \mathbb{R}$ be $L$-Lipschitz. Then
$$ \mathbf{Var}(f(X)) \leq L^2 \mathbf{Var}(X). $$
Proof. We adopt the symmetrization trick. Let $X'$ be an i.i.d. copy of $X$. Then
\begin{align*}
\mathbf{Var}(f(X))
= \frac{1}{2}\mathbf{E}[(f(X) - f(X'))^2]
\leq \frac{1}{2} \mathbf{E}[L^2 (X - X')^2]
= L^2 \mathbf{Var}(X).
\end{align*}

So by the Cauchy–Schwarz inequality and the above lemma,
\begin{align*}
\mathbf{Cov}(\widetilde{Y}, \widetilde{Z})
\leq \sqrt{\mathbf{Var}(f(X))\mathbf{Var}(g(X))}
\leq \mathbf{Var}(X).
\end{align*}
Therefore OP's claim holds with $C = 1$.

Addendum. By the Harris inequality and the monotonicity of $f$ and $g$, we also have
$$ \mathbf{Cov}(\widetilde{Y}, \widetilde{Z}) \geq 0. $$
This shows that the magnitude of the covariance $\mathbf{Cov}(\widetilde{Y}, \widetilde{Z})$ is bounded by the variance of $X$.
