Inequalities  involving moments $\newcommand{\bR}{\mathbb{R}}$  Suppose that $w:\bR\to \bR$ is a nonnegative, even smooth function  decaying fast at $\infty$, $w\in\mathscr{S}(\bR)$.
Define
$$s_m(w)= \int_{\bR^m} w(|x|) dx,\;\;  d_m(w):=\int_{\bR^m} x_i^2 w(|x|) dx,\;\;\forall i $$
$$ h_m(w) = \int_{\bR^m} x_i^2x_j^2 w(|x|) dx,\;\;\forall i\neq j.  $$
Is it true that
$$ d_m(w)^2\geq s_m(w) h_m(w) \tag{A} $$
for any $m\geq 2$ and any $w$  satisfying the above restrictions?
Where do the quantities $s_m,d_m,h_m$ come from?  
Consider a  smooth compact $m$-dimensional Riemann manifold $(M,g)$. Fix an orthonormal basis of $L^2(M)$ consisting of eigenfunctions $\Psi_k$
$$\Delta \Psi_k=\lambda_k \Psi_k, $$
$$0=\lambda_0<\lambda_1\leq \lambda_2 \leq \cdots $$
$\newcommand{\ve}{{\varepsilon}}$ For $\ve>0$ and $w$ as above we set $w_\ve(t):=w(\ve t)$. Consider random functions on $M$ of the type
$$ u_\ve =\sum_{k\geq 0} u_k \sqrt{w_\ve(\lambda_k^{1/2})} \Psi_k, $$
where the $u_k$ are independent normal r.v. with mean $0$ and  variance $1$. Note that thhe variance of  $u_k \sqrt{w_\ve(\lambda_k^{1/2})}$ is $w_\ve(\lambda_k^{1/2})$ which goes to zero very fast due to the fast decay of $w$. This  guarantees that $u_\ve$ is  almost surely smooth.  $\newcommand{\bp}{\boldsymbol{p}}$ Fix a point $\bp\in M$ and normal  coordinates $(x^i)$ at $\bp$. $\newcommand{\pa}{\partial}$ The numbers $s_m(w)$ $d_m(w)$, $h_m(w)$,  are related to the behavior  of the  random  variables $u_\ve(\bp)$, $\pa_{x^i}u_\ve(\bp)$ and $\pa^2_{x^ix^j}u_\ve(\bp)$ as $\ve\to 0$.  More precisely the rescaled variables $\ve^mu_\ve(\bp)$, $\ve^{m+1}\pa_{x^i}u_\ve(\bp)$ and $\ve^{m+2}\pa^2_{x^ix^j}u_\ve(\bp)$, $i\neq j$,   converge as $\ve \to 0$ to mean zero normal variables of variances $s_m(w)$, $d_m(w)$ and respectively $h_m(w)$.
An Example (showing that Jochen Wengenroth's example is not a counterexample.)  Observe first that
$$ s_n(w)= \left(\int_{S^{m-1}} dA\right)\int_0^\infty r^{m-1} w(r) dr $$
$$ d_n(w)= \left( \int_{S^{m-1}}x_1^2 dA(x)\right)\int_0^\infty r^{m+1} w(r) dr, $$
$$ h_n(w)= \left( \int_{S^{m-1}}x_1^2x_2^2 dA(x) \right)\int_0^\infty r^{m+3} w(r) dr, $$
and
$$a_m:=\int_{S^{m-1}} dA = \frac{2\pi^{\frac{m}{2}}}{\Gamma(\frac{m}{2})},\;\;  b_m:=\int_{S^{m-1}}x_1^2 dA(x)=   \frac{\pi^{\frac{m}{2}}}{\Gamma(1+\frac{m}{2})}=\frac{a_m}{m}, $$
$$c_m:= \int_{S^{m-1}}x_1^2x_2^2 dA(x)=   \frac{\pi^{\frac{m}{2}}}{2\Gamma(2+\frac{m}{2})} = \frac{b_m}{m+2}=\frac{a_m}{m(m+2)}. $$
Thus
$$ d_m^2= b_m^2 \left(\int_0^\infty r^{m+1} w(r) dr)\right)^2=\frac{a_m^2}{m^2}   \left(\int_0^\infty r^{m+1} w(r) dr)\right)^2, $$
and
$$s_m h_m= \frac{a_m^2}{m(m+2)} \left(\int_0^\infty r^{m-1} w(r) dr\right)\left(\int_0^\infty r^{m+3} w(r) dr\right), $$
so that the  inequality (A) is equivalent to 
$$ \left(\int_0^\infty r^{m+1} w(r) dr)\right)^2\geq \frac{m}{m+2} \left(\int_0^\infty r^{m-1} w(r) dr\right)\left(\int_0^\infty r^{m+3} w(r) dr\right). \tag{B}$$ 
Let us now choose $w(t)=t^{2k} e^{-t^2}$, $k$ nonnegative integer. Then for any $a>0$ we have
$$ \int_0^\infty t^a w(t) dt=\int_0^\infty t^{a+2k} e^{-t^2}  dt $$
($s=t^2$)
$$= \frac{1}{2}\int_0^\infty s^{\frac{a+2k-1}{2}} e^{-s} ds = \frac{1}{2}\Gamma\left(k+\frac{a+1}{2}\right). $$
For this choice of weight   the inequality becomes
$$\Gamma(k +1+\frac{m}{2})^2\geq \frac{m}{m+2}\Gamma(k+\frac{m}{2})\Gamma(k+2+\frac{m}{2}). $$
This is equivalent to 
$$ k+\frac{m}{2}=\frac{\Gamma(k +1+\frac{m}{2})}{\Gamma(k+\frac{m}{2})}\geq \frac{m}{m+2} \frac{\Gamma(k+2+\frac{m}{2})}{\Gamma(k+1+\frac{m}{2})}= \frac{m}{m+2}\left(k+1+\frac{m}{2}\right) . $$
One can easily verify that the last inequality  holds for any $m\geq 2$, $k\geq 0$. 
Update 1. The inequality (A) is true for weights $w$ of the form $w(t)=(1+t^2)e^{-t^2}$ and $w(t)=t^{2k}e^{-t^2}$. 
Update 2.   As Mikael de la Salle indicated  the inequality (A) is not valid in the   stated generality.      I want to rephrase the question: for which weights $w$ one has
$$\lim_{m\to\infty} \frac{d_m^2}{s_mh_m}=1,$$
i.e.,
$$ \lim_{m\to\infty}\frac{\left(\int_0^\infty r^{m+1}w(r) dr\right)^2}{\left(\int_0^\infty r^{m-1}w(r) dr\right)\left(\int_0^\infty r^{m+3}w(r) dr\right)}=1.\tag{C} $$
I could not find weights $w$ violating (C). That does not mean that there aren't any.  
Here is a geometric interpretation of  (C). Denote by $(-,-)_w$ the $L^2$-inner product    with respect to the measure $w(r)dr$ on $(0,\infty)$. We denote by $\Vert-\Vert_w$ the associated norm. If we   set 
$$\mu_k(r):=r^k,\;\;\nu_k(r):=\frac{1}{\Vert \mu_k \Vert_w} \mu_k(r),  $$
then the inequality (C) takes the form
$$ \lim_{m\to\infty} \bigl(\nu_{(m-1)/2}\;, \;\nu_{(m+3)/2}\bigr)_w=1.\;\;\tag{D} $$
This implies that,  as $m\to\infty$, the distance between the lines spanned by $\mu_{(m-1)/2}$ and $\mu_{(m+3)/2}$  goes to zero.
I'll set
$$I_k(w):=\int_0^\infty r^kw(r) dr. $$
Update 3. Mikael de la Salle strikes again! Following his suggestion   consider a weight $w$ such that 
$$ w(r)= \exp(-(\log r)\log(\log r)),\;\;\forall r\geq 1.$$
Then
$$ I_k(w)\sim J_k =\int_0^\infty t^k \exp(-(\log r)\log(\log r)) dr,\;\;\mbox{as $t\to \infty$}. $$
The last integral   can be estimated   using  the Laplace method and yields
$$J_k\sim\sqrt{2\pi\tau _k}  e^{\tau_k}, \;\;\tau_k=e^{k+1}$$
In particular this shows that for this particular weight one has
$$\lim_{m\to\infty} \frac{I_{m+1}(w)^2}{I_{m-1}(w)I_{m+3}(w)}=0. $$
Mikael's
 suspicions were right.
 A: I don't think so: To write things probabilistically let us assume $m=2$ and $s_2=1$ (the inequality is homogeneous w.r.t. the weight $w$). Then $w(|x|)$ is the Lebesgue-density of a
probability measure on $\mathbb R^2$ which is invariant under rotation (since it only depends on the norm) and for the two coordinate functions $X,Y$ you are asking for
$$ E(X^2 Y^2) \le E(X^2) E(Y^2).$$
(For the special weight $w_0(t)=c\exp(-t^2/2)$ the coordinates are  independent and one has even equality.) 
In the general case, the random vector $2^{-1/2}(X+Y,X-Y)$ is a rotation of $(X,Y)$ and
has thus the same distribution as $(X,Y)$ which leads to 
$$ E(X^2 Y^2)= \frac{1}{4} E((X+Y)^2(X-Y)^2) = \frac{1}{4} E((X^2-Y^2)^2) = \frac{1}{2} (E(X^4)-E(X^2Y^2))$$ hence $E(X^2Y^2)= \frac{1}{3} E(X^4)$.
Now, if the "tails" of $w$ are "heavier" than those of $w_0$ one gets that the
inequality might be wrong. 
A: As explained by Jochen Wengenroth, the answer is no in general.
Indeed, as you notice, the inequality $(A)$ is equivalent to an inequality about the moments $B_k=\int_0^\infty t^k w(t) dt$, namely $B_{m+1}^2 \geq C(m) B_{m-1} B_{m+3}$ for some $C(m)$ which is bounded above and below (my rapid computations give $C(m)=(1+2/m)^{-1}$ and do not agree with yours, so I prefer to be vague with the exact value of $C(m)$).
Anyway, for any $m$ you can find $w$ such that $B_{m+1}^2/(B_{m-1} B_{m+3})$ is arbitrarily close to zero. Indeed, if take $\mu_n$ the measure on $\mathbb R$ giving mass $1$ to $1$ and mass $1/n^{m+2}$ to $n$, then $ \int t^k d\mu_n = 1+n^{k-m-2}$, so that the ratio $B_{m+1}^2/(B_{m-1} B_{m+3})$ is equivalent to $1/n$, which goes to $0$. Of course, $\mu_n$ is not of the form $w(r)dr$, but it can be approximated by such measures.
But the reverse inequality is almost true, namely by Hoelder's inequality $B_{m+1}^2\leq(B_{m-1} B_{m+3})$, so that $d_m(w)^2/(s_m(w) h_m(w))$ is bounded above by $C(m)$. 
