Which inequality for the fourth central moment do we need to apply here? Let


*

*$d\in\mathbb N$ with $d>1$

*$\lambda^d$ denote the Lebesuge measure on $\mathcal B\left(\mathbb R^d\right)$

*$f\in C^2(\mathbb R)$ be positive and $$\pi(x):=\prod_{i=1}^df(x_i)\;\;\;\text{for }x\in\mathbb R^d$$

*$(\Omega,\mathcal A,\operatorname P)$ be a probability space

*$X:\Omega\to\mathbb R^d$ with $X_\ast\operatorname P=\pi\lambda^d$
Now, let $$g(x):=\frac1{d-1}\sum_{i=2}^d\left|\frac{f'(x_i)}{f(x_i)}\right|^2\;\;\;\text{for }x\in\mathbb R^d.$$
Assume $$M:=\int\frac{|f'|^8}{f^7}\:{\rm d}\lambda^1<\infty\tag1.$$
Note that $$\operatorname E\left[g(X)\right]=\int\frac{\left|f'\right|^2}f\:{\rm d}\lambda^1=:I.\tag2$$

I want to show that $$\operatorname E\left[\left|g(X)-I\right|^4\right]\le d^{-\frac12}(d-1)^{-\frac32}3M\tag3.$$

Is there an easy estimate which yields $(3)$? Clearly, we can expand the left-hand side using the multinomial theorem, but then we deal with a complicated expression and annoying computations.
On the other hand, by applying the Cauchy-Schwarz inequality twice, we obtain $$\operatorname E\left[\left|g(X)-I\right|^4\right]\le\frac1{d-1}\sum_{i=2}^d\operatorname E\left[\left|\left|\frac{f'(X)}{f(X)}\right|^2-I\right|^4\right],$$ but I don't know how we need to proceed from here.
 A: Let us write $h = (f'/f)^2$.
By the definitions of $I$ and $M$, $E[h(X)] = I$ and $E[h^4(X)] = M$,
and we have
$$ g(x) = \frac{1}{d-1} \sum_{i=2}^{d} h(x_i).$$
Now we have
\begin{align}
    E \left[ |g(X) - I|^4 \right] &= E \left[ \left( \frac{1}{d-1} \sum_{i=2}^{d} (h(X_i) - I) \right)^4 \right] \\
                                  &= \left( \frac{1}{d-1} \right)^4  E \left[ \left(\sum_{i=2}^{d}  (h(X_i) - I)^4 \right) + 3 \left( \sum_{i=2}^{d} \sum_{j=2, j\neq i}^{d} (h(X_i) - I)^2(h(X_j) - I)^2 \right) \right],
\end{align}
because the other terms of the product cancel out from the fact that $E(h(X_i) - I) = 0$. 
Using the moment bound at our disposal and Holder's inequality,
we obtain
\begin{align}
    E \left[ |g(X) - I|^4 \right] &\leq M \, \left( \frac{1}{d-1} \right)^4  \left(d - 1 + 3 (d-1)(d-2) \right), \\
                                  &= 3M \,  \frac{d-5/3}{(d-1)^3} = 3 M \, (d-1)^{-3/2} \, \frac{d - 5/3}{(d-1)^{3/2}}.
\end{align}
and the last term is less than $d^{-1/2}$ when $d \geq 2$.

EDIT: the fact that $E[(h(X_i) - I)^4] \leq  E[h(X_i)^4]$ follows from the non-negativity of $h$, see my answer here.
