Estimate for the trace of some self adjoint non negative trace class operators

Question

Let $T:L^2(\mathbb{R}^3)\to L^2(\mathbb{R}^3)$ be a self-adjoint trace-class operator and denote by $0\leq K(x,y)\leq 1$ its kernel with $K(x,y)$ regular and decaying sufficiently fast. The following is supposed to be an "easy to prove" inequality $$ \big\Vert \nabla \sqrt{K(\cdot,\cdot)}\big\Vert_{L^2}^2 \leq \text{tr}\big(\sqrt{-\Delta} \ T\sqrt{-\Delta}\big) $$ (assuming that the operator on the RHS is also trace-class), however, I don't really see how to approach it. I assumed that one should use a spectral decomposition for $T$, to deal with the RHS, but I don't see how one could obtain something similar or comparable to the LHS, mainly because of the derivative acting on the square root. I was thinking on writing the RHS as $$ \text{tr}\big(\sqrt{-\Delta} \ T\sqrt{-\Delta}\big) = \sum_n \big\langle e_n,\big(\sqrt{-\Delta} \ T\sqrt{-\Delta}\big) e_n\big\rangle, $$ but then I don't see how could I obtain a bound as the one on the LHS. Is that the correct approach or is there a better starting point?