Variance and covariance inequality Given a real-valued random variable $X$, is
$$2\mathbb E[X] \mathrm{Var}(X) \geq \mathrm{Cov}(X, X^2)$$
true?
Any pointers for how to tackle this problem would be immensely helpful.
 A: Use the variance and covariance identities
$$\text{Var}(X)=\mathbb E[X^2] − \mathbb E[X]^2$$
and
$$\text{Cov}(X,Y) = \mathbb E[XY] − \mathbb E[X] \mathbb E[Y]$$
Then the given inequality is equivalent to
$$ 2 \mathbb E[X] \Big( \mathbb E[X^2] - \mathbb E[X]^2 \Big)
\overset{?}{\geq}
\mathbb E[X^3] - \mathbb E[X] \mathbb E[X^2] \tag{1}$$
For a counterexample, let $X \sim \text{Exp}(\lambda)$ be the exponential distribution with parameter $\lambda \in (0, \infty)$ and probability mass function
$$ f_X(x) = \begin{cases}
\lambda e^{-\lambda x} & \text{if } x \geq 0 \\
0 & \text{if } x < 0
\end{cases} $$
Its $n$-th moment is given by $\mathbb E[X^n]=\frac{n!}{\lambda^n}$, so the inequality $(1)$ becomes
$$2 \cdot \frac{1}{\lambda} \Big( \frac{2}{\lambda^2} - \frac{1}{\lambda^2} \Big)
\geq
\frac{6}{\lambda^3} - \frac{1}{\lambda} \cdot \frac{2}{\lambda^2}$$
which reduces to
$$\frac{2}{\lambda^3} \geq \frac{4}{\lambda^3}$$
This is false for all $\lambda \in (0, \infty)$, hence we found a counterexample.
A: Alternate approach / observation...
Assume $2 E[X] Var(X) \ge Cov(X, X^2)$ for any real-valued r.v. $X.$  Then in particular it would also be true for $-X$, so that:
$$
\begin{align}
2E[-X] Var(-X) &\ge Cov(-X, (-X)^2)\\
-2E[X] Var(X) &\ge -Cov(X, X^2)\\
2E[X] Var(X) &\le Cov(X, X^2)
\end{align}
$$
which, combined with the original inequality, means: $2E[X]Var(X) = Cov(X, X^2)$ for all $X$.  
At this point, it would seem one should be able to find a counter-example easily, without using e.g. $Cov(X, X^2) = E[X^3] - E[X]E[X^2]$ etc, but I just couldn't think of a quick one.  :(
