Showing that $0\leq A\leq B$ and $B \in \mathcal{L}_c(H)$ implies that $A \in \mathcal{L}_c(H)$.

Solution:

Because $B-A \ge 0$, then $[x,y]=\langle (B-A)x,y\rangle$ is a pseudo inner product, lacking only positive definiteness. As such, the Cauchy-Schwarz inequality holds $$ |[x,y]|^2 \le [x,x][y,y] \\ |\langle (B-A)x,y\rangle|^2 \le \langle (B-A)x,x\rangle\langle (B-A)y,y\rangle. $$ Now set $y=(B-A)x$ in the above to obtain $$ \|(B-A)x\|^4 \le \langle (B-A)x,x\rangle\langle(B-A)(B-A)x,(B-A)x\rangle \\ \|(B-A)x\|^4 \le \langle (B-A)x,x\rangle\|B-A\|\|(B-A)x\|^2 \\ \|(B-A)x\|^2 \le \|B-A\|\langle(B-A)x,x\rangle \\ \|(B-A)x\|^2 \le \|B-A\|\langle Bx,x\rangle. $$ Suppose $\{ x_n \}$ is a bounded sequence. Because $B$ is compact, there is a subsequence $\{ x_{n_k} \}$ such that $\{ Bx_{n_k} \}$ converges. By the above, $\{ (B-A)x_{n_k}\}$ is a Cauchy sequence and, hence, converges to some $y$. But $\{ Bx_{n_k} \}$ also converges. Therefore $\{ Ax_{n_k} \}$ converges, leading to the conclusion that $A$ is compact.