$\newcommand{\bR}{\mathbb{R}}$ 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.
