This question popped up to me while trying to solve an other exercise, but now I got interested into the question by itself.
Let's take a positive r.v. $X$.
Is there a way to bound with a constant $K$, independent of $X$, the following product:
$$E\left[ \frac{1}{(1+X)^2} \right ]E[X]$$
I conjecture such a constant may exist since when $X$ is large $\frac{1}{(1+X)^2}$ is small (close to zero) and the contributions could balance.
Random trials: I tried to use that $E\left[ \frac{1}{(1+X)^2} \right]=\int_0^{+\infty}P\left(\frac{1}{(1+X)^2}>t\right)dt$ and the Markov inequality to link the integrand with $E[X]$, but without success.
Replacing the expectations with integrals w.r.t. a probability density I think one could pose the question as a constrained calculus of variations optimization (this could also give the optimal constant) but I was not able to finish this approach (not so used to that formalism).
Consider this example:
Then $E[X_n]>\dfrac12 n$ and $E\left[\dfrac{1}{(1+X_n)^2}\right]>\dfrac12 \dfrac{1}{(1+1)^2} = \frac18$
so $E\left[\dfrac{1}{(1+X_n)^2}\right]E[X_n]>\dfrac 1{16}n$, which increases without limit as $n$ increases.
Any example with a positive lower bound on $E\left[\frac{1}{(1+X)^2}\right]$ and no upper bound on $E[X]$ would do