Trying to bound one functional by another functional In my little research project, I faced the following problem: Assume that $\rho$ is a probability density function with support $[0,\infty)$ and 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!
Edit: taking into account of the answer written by Iosif Pinelis, there is no hope for a non-trivial $f$ if we do not post further restrictions on the pdf $\rho$, I am wondering if we can hope for a non-trivial $f$ (meaning $f$ is not identically zero) if we only look at "smooth" $\rho$.
 A: Such a function $f$ must be identically zero.
Indeed, by approximation, the problem can be restated as follows:

Suppose that $f\colon[0,\infty)\to[0,\infty)$ is a function such that for any nonnegative iid random variables (r.v.'s) $X,Y,Z$ we have
$$H\ge f(G),\tag{1}$$
where
$$G:=E|X-Y|,\quad H:=E|X-Y|-\tfrac12\,E|X+Y-2Z|.$$
Prove that then $f=0$.

To prove this, take any real $a\ge0$ and let each of the r.v.'s $X,Y,Z$ take each of the values $0,2a$ with probability $1/2$, if $a>0$; if $a=0$, let each of the r.v.'s $X,Y,Z$ take the only value $0$.
Then $G=a$ and $H=0$, so that (1) implies $0\ge f(a)$, for each real $a\ge0$. Since $f\ge0$, we conclude that $f=0$.
$\Box$
Remark 1: Here, more generally, instead of assigning probability $1/2$ to each of the values $0,2a$, we can assign probabilities $p$ and $q=1-p$ to the values $0$ and $a/(2pq)$, respectively, for any $p\in(0,1)$.
Response to a comment by the OP: The OP is saying in that comment that the mean $\mu$ was assumed to be fixed. That assumption was not stated in the OP. It was only said that $\mu>0$. However, even assuming $\mu$ is fixed, Remark 1 above gives essentially the same conclusion, that $f=0$. Indeed, in the conditions of Remark 1, for any $a\in(0,2\mu)$, take $p=a/(2\mu)\in(0,1)$. Then $EX=a/(2p)=\mu$, as needed, and $f(a)=0$. So, $f=0$ on the interval $(0,2\mu)$ and hence, in view of the condition $f(0)=0$, we have $f=0$ on the interval $[0,2\mu)$.
Note also that we only need to define $f$ on the interval $[0,2\mu)$, because $E|X-Y|\in[0,2\mu)$. Indeed, by the triangle inequality, $E|X-Y|\le EX+EY=2\mu$, and $E|X-Y|=EX+EY$ only if $|X-Y|=X+Y$ almost surely (a.s.), which may happen only if $X=0$ a.s., which would imply $\mu=0$, which would contradict the assumption $\mu>0$.
