We know the following relationship: If A and B are positive semidefinite matrices, then $\mathrm{tr}(AB)^2\le\mathrm{tr}(A^2B^2)$.
How can I prove this relationship? If A and B are positive semidefinite matrices, then $$\mathrm{tr}(A^3B + AB^3)\ge 2 \mathrm{tr}( A^2B^2).$$
In fact this follows easily from that the matrices are diagonalizable in an orthogonal basis , have non-negative eigenvalues, and that $x^3y+xy^3- 2y^2x^2\ge 0$ for $x\ge 0,y\ge 0$.
$$A = \sum_{i=1}^n a_iu_i u_i^\top,\qquad B=\sum_{j=1}^n b_jv_j v_j^\top$$ So $$A^k = \sum_{i=1}^n a_i^ku_i u_i^\top,\qquad B^m=\sum_{j=1}^n b_j^mv_j v_j^\top$$ and $$Tr(A^3B+AB^3-2A^2B^2) = \sum_{i=1}^n \sum_{j=1}^n (a_i^3b_j+a_i b_j^3-2a_i^2 b_j^2) Tr(u_i u_i^\top v_j v_j^\top)$$
Conclude with $Tr(u_i u_i^\top v_j v_j^\top)=(u_i^\top v_j)^2$.