possibility of bounding one functional by another functional This is a natural follow-up question related to one of my previous questions at here. Assume that $\rho$ is a log-concave probability density function with support $[0,\infty)$ and fixed mean $\mu >0$. Let $$H[\rho] = \iiint_{y,v,w\geq 0} \rho(v)\rho(w)\rho(y)\left(\frac{|v-y|+|w-y|}{2} - \left|\frac{v+w}{2} -y\right|\right)\,\mathrm{d}v\,\mathrm{d}w\,\mathrm{d}y \,\,(\geq 0) $$ and $$G[\rho] = \iint_{x,y\geq 0} \rho(x)\rho(y)|x-y|\,\mathrm{d}x\,\mathrm{d}y.$$ I am wondering if it is possible to obtain functional inequalities of the form $$H[\rho] \geq f(G[\rho])$$ for some non-negative function $f \colon [0,\infty) \to [0,\infty)$ with $f(0) = 0$. Of course, the best scenario I can hope for is for $f(x) = c\cdot x$ for some $c >0$, but this is might be too good to be true. Thanks for your help!

Remark: taking into account of the comment made by Iosif Pinelis, there is no hope for a non-trivial $f$ if we do not impose any "structural" restrictions on the pdf $\rho$ (such as log-concavity of the density).
 A: $\newcommand{\R}{\mathbb R}$Let $p:=\rho$, $H:=H[p]$, $G:=G[p]$. Let us show that
\begin{equation*}
    f(x)\equiv kx
\end{equation*}
with
\begin{equation*}
    k=1/14334
\end{equation*}
will do. We will not need the restriction that the support of the distribution is $[0,\infty)$.
By approximation, without loss of generality (wlog), $p(x)>0$ for all real $x$ -- for instance, one may approximate $p$ by its convolution $p*g$ with the density $g$ of a centered normal distribution with an arbitrarily small variance. Then $p*g>0$ on $\R$ and $p*g$ is arbitrarily close to $p$ and log concave, since $p$ and $g$ are log concave -- see e.g. this Wikipedia article.
Therefore and because $p$ is a log-concave density, $p$ is continuous and attains its maximum value, say $p_*[>0]$, at some point $c\in\R$, so that $p_*=p(c)\ge p(x)$ for all real $x$. Moreover, again because $p$ is a log-concave density, there exist (unique) real $a$ and $b$ such that
\begin{equation*}
\text{$a<c<b$ and $p(a)=p(b)=p_*/e$. }  
\end{equation*}
Again by the log-concavity of $p$,
\begin{equation*}
    p(x)\le q(x):=
    \left\{
    \begin{aligned}
    q_1(x):=p_*\exp\Big\{-\frac{x-c}{a-c}\Big\} &\text{ if }x<a, \\ 
    p_* &\text{ if }a\le x<b, \\ 
    q_2(x):=p_*\exp\Big\{-\frac{x-c}{b-c}\Big\} &\text{ if }x\ge b. 
    \end{aligned}
    \right.
\end{equation*}
As an illustration, for $p(x)=xe^{-x}1(x>0)$, here are the graphs $\{(x,p(x))\colon-2\le x\le6\}$ (blue), $\{(x,q(x))\colon-2\le x\le6\}$ (black), $\{(x,q_1(x))\colon a\le x\le c\}$ (dashed), and $\{(x,q_2(x))\colon c\le x\le b\}$ (dashed):

For this particular $p$, we have $c=1$, $a=-W_0\left(-1/e^2\right)=0.15859\dots$, and $b=-W_{-1}\left(-1/e^2\right)=3.1461\dots$, where $W_j$ is the $j$th branch of the Lambert $W$ function.
By shifting, wlog
$$a=0.$$
So,
\begin{align*}
    G&\le\iint\limits_{\R^2} q(x)q(y)|x-y|\,dx\,dy \\ 
    &=p_*^2\frac{e^2 b^3+9 e b^3+3 b^3+3 b^2 c-12 e b^2 c-3 b c^2+12 e b c^2}{3 e^2} \\ 
    &\le p_*^2\frac{\left(1+3 e+e^2/3\right) b^3}{e^2},
\end{align*}
since $0<c<b$.
Moreover, again by the log-concavity of $p$, we have $p\ge p_*/e$ on the interval $[a,b]=[0,b]$, so that $1=\int_\R p\ge\int_0^b p_*/e=bp_*/e$, whence $p_*\le e/b$ and
\begin{equation*}
    G\le (1+3 e+e^2/3) b. \tag{1}
\end{equation*}
On the other hand, because $p\ge p_*/e$ on the interval $[a,b]=[0,b]$ and the integrand in the definition of $H$ is $\ge0$, we have
\begin{align*}
    H&\ge\Big(\frac{p_*}e\Big)^3\iiint\limits_{[0,b]^3} \Big(\frac{|x-z|+|y-z|}2
    -\Big|\frac{x-z+y-z}2\Big|\Big)\,dx\,dy\,dz \\
    &=\Big(\frac{p_*}e\Big)^3\frac{b^4}{24}. 
\end{align*}
Also, $1=\int_\R p\le\int_\R q=p_*b(1+1/e)$, so that $p_*\ge1/(b(1+1/e))$ and hence
\begin{equation*}
    H\ge\Big(\frac1{(e+1)b}\Big)^3\frac{b^4}{24}=\frac b{24(e+1)^3}. \tag{2}
\end{equation*}
Comparing (1) and (2), we get
\begin{equation*}
    H\ge\frac G{24(e+1)^3(1+3 e+e^2/3)}\ge\frac G{14334},
\end{equation*}
as claimed.
