Reverse trace inequality for powers greater than one

Let $A,B\geq 0$. Let $f$ be a nonnegative and nondecreasing function on an interval $\mathcal{J}$ such that $\operatorname{spec}(A)\subseteq\mathcal{J}$. Reverse trace inequality. One has

$$\operatorname{Tr}\left[ f(A) A^s B^s \right] \geq \operatorname{Tr}\left[ f(A) \left(A^{1/2} B A^{1/2} \right)^s \right], \quad \forall\,s\geq 1.$$

This conjectures the reverse of the proved trace inequality for $s\in[0,1]$. The authors note that operator convexity of $x\mapsto x^s$ may establish it for $s\in[1,2]$ under a suitable relaxation of an auxiliary theorem's hypotheses; the full range $s\geq 1$ remains open.