We conjecture that a similar bound holds without the parameter ${\varepsilon}$ in dimension $d=2,3$ as well.

In dimension $d=3$ one can take $$\kappa_1=1,\qquad \kappa_2=48 \qquad\text{in~\eqref{eq:upper_bound_T_kappa}.}$$ These constants are not optimal and they are only displayed for concreteness. Our proof allows to slightly improve the constants under the assumption that ${\varepsilon}$ is small enough. For instance for ${\varepsilon}{\leqslant}1$, we have the better inequality $$T(\rho){\leqslant}q^{-\frac23}c_{\rm TF}\left(1+\frac{{\varepsilon}}{15}\right)\int_{{\mathbb{R}}^3}\rho(x)^{\frac53}dx
 +\frac{19}{{\varepsilon}}\int_{{\mathbb{R}}^3}|\nabla\sqrt{\rho}(x)|^2\,dx.
 \label{eq:upper_bound_T_eps}$$

For simplicity we only write the proof in the no-spin case $q=1$. Recall that the free Fermi sea $$P_r={{\ensuremath {\mathds 1} }}\left(-\Delta{\leqslant}\frac{d+2}d c_{\rm TF}r^{2/d}\right)$$ has the constant density $\rho_{P_r}=r$ and the constant kinetic energy density $c_{\rm TF}r^{1+2/d}$. In particular, if we take $$\gamma=\sqrt{f(x)}\,{{\ensuremath {\mathds 1} }}\left(-\Delta{\leqslant}\frac{d+2}dc_{\rm TF}r^{2/d}\right)\sqrt{f(x)}$$ for some $f{\geqslant}0$, we obtain $$\rho_\gamma=rf(x),\qquad {{\rm Tr}\,}(-\Delta)\gamma=r\int_{{\mathbb{R}}^d}|\nabla \sqrt{f}(x)|^2\,dx+c_{\rm TF}r^{1+\frac2d}\,\int_{{\mathbb{R}}^d}f(x)\,dx.
\label{eq:computation_simple_trial_state}$$ See for instance [@LewLieSei-18 Sec. 5] for details. In addition, we have in the sense of operators $0{\leqslant}\gamma{\leqslant}f(x)$, hence $\gamma$ is a fermionic one-particle density matrix under the additional condition that $f{\leqslant}1$.

Let now $\eta{\geqslant}0$ be a smooth non-negative function such that $$\int_0^{\infty}\eta(t)\,dt=1,\qquad \int_0^{\infty}\eta(t)\,\frac{dt}{t}{\leqslant}1.
\label{eq:cond_eta}$$ Using the smooth layer cake principle $$\rho(x)=\int_0^{\infty}\eta\left(\frac{t}{\rho(x)}\right)\,dt$$ we introduce the trial state $$\gamma=\int_0^{\infty}\sqrt{\eta\left(\frac{t}{\rho(x)}\right)}\;{{\ensuremath {\mathds 1} }}\left(-\Delta{\leqslant}\frac{d+2}d c_{\rm TF}\,t^{2/d}\right)\;\sqrt{\eta\left(\frac{t}{\rho(x)}\right)}\;\frac{dt}{t}.$$ In the sense of operators, we have $$0{\leqslant}\gamma{\leqslant}\int_0^{\infty}\eta\left(\frac{t}{\rho(x)}\right)\;\frac{dt}{t}=\int_0^{\infty}\eta(t)\,\frac{dt}{t}{\leqslant}1.$$ In addition, $\gamma$ has the 