EDIT: I have changed the question to have less parameters, fitting it into the context of Orlicz spaces.
Suppose $f:[0,\infty)\to[0,\infty)$ is concave convex and increasing, $g:[0,\infty)\to[0,\infty)$ f^{-1}:[0,\infty)\to[0,\infty)$ is convex concave and increasing , and that $f\circ g$ is convex. Then by Jensen's inequality we know that
$f\circ g\left( \int_{B(0,\delta)} |\nabla u(z)| \rho^\delta(z)dz \right) \leq \int_{B(0,\delta)}f\circ g\left( |\nabla u(z)|\right) \rho^\delta(z)dz $
whenever $\int_{B(0,\delta)} \rho^\delta(z)dz =1$.
Is it true that
$f\left(\int_{B(0,h)} g\left(\int_{B(0,\delta)f^{-1}\left(\int_{B(0,h)} f\left(\int_{B(0,\delta)} |\nabla u(x-z)| u(x,z)| \rho^\delta(z)dz\right)\;dx\right)\leq \int_{B(0,\delta)} f\left(\int_{B(0,h)f^{-1}\left(\int_{B(0,h)} g\left(|\nabla u(x-z)| f\left(|u(x,z)| \right)\;dx\right)\rho^\delta(z)dz$?
A partial result is that Jensen's inequality for $f$ and Tonelli's Theorem implies $f\left(\int_{B(0,h)} g\left(\int_{B(0,\delta)} |\nabla u(x-z)| \rho^\delta(z)dz\right)\;dx\right)\leq f\left( \int_{B(0,\delta)} \int_{B(0,h)} g\left(|\nabla u(x-z)| \right)\;dx \rho^\delta(z)dz\right)$,
which is as far as we can go, because $f$ is concave.
When $f(t)=t^\frac{1}{p}$ and f(t)=t^p$, then $g(t)=t^p$, f^{-1}(t)=t^\frac{1}{p}$ and this can be seen through is precisely Minkowski's Inequality for integrals, as the title suggests. I am wondering if it holds under these more general conditions. Proof or reference, and the proof uses duality in $L^p$. Does this theorem stretch to some class of $N$-functions $f$ ($\Delta_2$, etc.)?
