For bounded operators $A,B$ on a Hilbert space $\mathbb{H}$ prove that for any $c \geq 0$, $A^*A \leq cB^*B \Leftrightarrow \lVert A \rVert \leq \sqrt{c}\lVert B \rVert$.
This is an exercise from Wolfgang Sherer's book, Mathematics of Quantum Computing. Here $A^*$ denotes the adjoint operator of $A$, $\lVert A \rVert$ denotes the operator norm and for operators $A, B$ the statement $A \leq B$ is defined as $B-A \geq 0$ (i.e for all $|\psi\rangle \in \mathbb{H}$, $\langle \psi | (B-A)\psi\rangle \geq 0$).
My work so far: Using the definitions $A^*A \leq cB^*B$ is equivalent to $\langle \psi | (cB^*B-A^*A)\psi\rangle \geq 0$ for all $|\psi\rangle \in \mathbb{H}$. Expanding the inner product this is equivalent to $\lVert A\psi\rVert \leq \sqrt{c}\lVert B\psi\rVert$. Now taking supremums over all $|\psi\rangle$ with $\lVert \psi \rVert = 1$ we get the $\Rightarrow$ direction.
My question: How can I proceed with the $\Leftarrow$ direction? In the solutions on the back of the book it states that $\lVert A\psi\rVert \leq \sqrt{c}\lVert B\psi\rVert \Leftrightarrow \lVert A \rVert \leq \sqrt{c}\lVert B \rVert$ citing equation $(2.57)$ which is found later in the book and is just the definition of trace of an operator. I don't think that this is correct though.