If $\operatorname{tr}(A+B)>\operatorname{tr}(A)$, does it hold that $\operatorname{tr}((A+B)^k)>\operatorname{tr}(A^k)$ for all $k\geq 1$ I wonder whether the following holds and if so how it could be proved:
Let $A, B$ be (non-commuting) positive semi-definite matrices,
If $\operatorname{tr}(A+B)>\operatorname{tr}(A)$, does it hold that $\operatorname{tr}((A+B)^k)>\operatorname{tr}(A^k)$ for all $k\geq 1$.
Any ideas would be welcome. Thank you very much in advance

Very partial answer:
It is true for k=2:
$tr((A+B)^2)=tr(A^2+AB+BA+B^2)=tr(A^2)+tr(AB)+tr(BA)+tr(B^2)=tr(A^2)+2tr(AB)+tr(B^2)$
by linearity of trace and by the fact that $tr(AB)=tr(BA)$. Furthermore, as B is psd, $tr(B^2)\geq 0$, and $tr(AB)\geq 0$ (see A.U. Kennington, Power concavity and boundary value problems, Indiana University Mathematics Journal Vol.34, No. 3, 1985, p. 687-704, Appendix). Thus $tr((A+B)^2)\geq tr(A^2)$
