$\newcommand{\e}{\operatorname E}$"Obviously" if $g$ is a weakly increasing function and $X$ and $g(X)$ are both random variables with finite variance, then the covariance (and hence the correlation) between $X$ and $g(X)$ is non-negative. But there is the question of how to prove this "obvious" proposition. My question is whether something simpler than what I wrote should be used instead of what I wrote.

\begin{align} \operatorname{cov}(X,g(X)) = {} & \e\Big( \big(X-\e(X)\big)\big(g(X) - \e(g(X))\big) \Big) \\[6pt] = {} & \e\Big( \big(X-\e(X)\big) \big( (g(X) - g(\e(X))) + (g(\e(X)) - \e(g(X)))\big) \Big) \\[6pt] = {} & \e\Big(\big(X-\e(X)\big)\big( g(X) - g(\e(X))\big)\Big) \\ & + \e\Big( \big( X-\e(X)\big)\big( \underbrace{g(\e(X)) - g(\e(X))} \big) \Big) \end{align}

Now an essential point is that the expression over the $\underbrace{\text{underbrace}}$ is constant, i.e. not random. Therefore it can be pulled out of the outermost expectation operator just after the "plus" sign. Then we're left with $$ \e\big( X-\e(X)\big) $$ and that is zero. In the first term we now have the expected value of $$ \big( X-\e(X)\big)\big( g(X) - g(\e(X))\big) $$ and that random variable is nonnegative with probability $1.$

This leaves me feeling that I may have missed a simpler proof, just because the proposition stated just after the scare-quoted "Obviously" above seems obvious.

Did I miss a simpler proof?

$\newcommand{\e}{\operatorname E}$"Obviously" if $g$ is a weakly increasing function and $X$ and $g(X)$ are both random variables with finite variance, then the covariance (and hence the correlation) between $X$ and $g(X)$ is non-negative. But there is the question of how to prove this "obvious" proposition. My question is whether something simpler than what I wrote should be used instead of what I wrote.

\begin{align} \operatorname{cov}(X,g(X)) = {} & \e\Big( \big(X-\e(X)\big)\big(g(X) - \e(g(X))\big) \Big) \\[6pt] = {} & \e\Big( \big(X-\e(X)\big) \big( (g(X) - g(\e(X))) + (g(\e(X)) - \e(g(X)))\big) \Big) \\[6pt] = {} & \e\Big(\big(X-\e(X)\big)\big( g(X) - g(\e(X))\big)\Big) \\ & + \e\Big( \big( X-\e(X)\big)\big( \underbrace{g(\e(X)) - \e(g(X)} \big) \Big) \end{align}

Now an essential point is that the expression over the $\underbrace{\text{underbrace}}$ is constant, i.e. not random. Therefore it can be pulled out of the outermost expectation operator just after the "plus" sign. Then we're left with $$ \e\big( X-\e(X)\big) $$ and that is zero. In the first term we now have the expected value of $$ \big( X-\e(X)\big)\big( g(X) - g(\e(X))\big) $$ and that random variable is nonnegative with probability $1.$

This leaves me feeling that I may have missed a simpler proof, just because the proposition stated just after the scare-quoted "Obviously" above seems obvious.

Did I miss a simpler proof?