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



Inequalities  involving moments
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 $C(m)$. 


