All roots $\lambda$ of $\det(A-\lambda B)=0$ are $\ge1$ when $B$ is p.d and $A-B$ is n.n.d. I am trying to prove the following statement:

Let $A,B\in M(n,\mathbb{R})$.
If $B$ is positive definite and $(A-B)$ is non-negative definite, then $\det(A-\lambda B)=0$ has all its roots $\lambda\geqslant1$ and conversely, if all roots $\lambda\geqslant 1$, then $(A-B)$ is non-negative definite.

If $(A-B)$ is n.n.d and $B$ is p.d, then I have $x^\top(A-B)x\geqslant0$ for all $x\in\mathbb{R}^n$.
This implies $x^\top Ax\geqslant x^\top Bx>0$ for all $x\ne0$, so that $A$ is p.d.
Moreover, as $B$ is p.d, $B$ is nonsingular.
So, $\det(A-\lambda B)=0\implies\det((AB^{-1}-\lambda I)B)=0$
$\qquad\qquad\qquad\qquad\quad\implies\det(AB^{-1}-\lambda I)=0$ , as $\det(B)\ne0$
Thus $\lambda$ is an eigenvalue of the matrix $AB^{-1}$.
Now for the eigenvector $x\ne0$ corresponding to $\lambda$ we have,
$(AB^{-1})x=\lambda x\implies(AB^{-1}-I)x=(\lambda-1)x$.
If I can show that $AB^{-1}-I$ is n.n.d given that both $A$ and $B$ are p.d, then that would possibly
imply $\lambda-1\geqslant0$ and I am done. But I am not sure if this is true or not.
In a different approach using the fact that a p.d matrix can be expressed as $D^\top D$ for some nonsingular matrix $D$, I was able to show that $AB^{-1}-I=PQ$ for some n.n.d matrix $P$ and p.d matrix $Q$. Does that help me conclude that $AB^{-1}-I$ is indeed n.n.d?
Any simpler or alternate approach is welcome.

You can finish your own proof with the expression $QD$ using this answer. Since $Q$ is p.d., $QD$ has the same eigenvalues as $Q^{1/2}DQ^{1/2}$. Since $x^TQ^{1/2}DQ^{1/2} x = (Q^{1/2}x)^TD(Q^{1/2}x) \geq 0$ (because $D$ is n.n.d.), we get that $QD$ is indeed n.n.d.
