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Plevnik (2016) (pdf) found the counterexample $$ A = \begin{bmatrix}76&0&0\\0&0&0\\0&0&1\end{bmatrix} \qquad\text{and}\qquad B = \begin{bmatrix}20&-14&13\\-14&2880&3100\\13&3100&3380\end{bmatrix} $$ with $$ \operatorname{tr} A^4BAB^4 = 7608677695167720100 > 7566365725138281700 = \operatorname{tr} A^5B^5. $$ This does not answer your question since $A$ is not positive definite, but unless I miscalculated, replacing entry $A_{22}$ by $0.01$ gives a pair that does: $$ \operatorname{tr} A^4BAB^4 = 7.58468087608\times10^{18} > 7.56636572557\times10^{18} = \operatorname{tr} A^5B^5. $$