Below is an exercise which I am trying to solve. The exercise is from Kadison and Ringrose's "Fundamentals of the Theory of Operator Algebra Volume I".
Let $\mathscr{H}$ be a Hilbert space and $K \in B(\mathscr{H})^+$ (set of bounded, positive operators). Prove: $$||K|| = \min\{a:a\in R,\ K \leq aI\}. $$
I have already shown that the equation $\langle x, y \rangle _1 = \langle Kx, y\rangle$ defines an inner product $\langle \cdot \ , \ \cdot \rangle _1$ on $\mathscr{H}$. I feel like I am suppose to use the Cauchy-Schwarz inequality for $\langle \cdot \ , \ \cdot \rangle _1$, but not sure how to apply it in this case.
My attempt so far:
$\langle (aI - K)x, x\rangle = a\langle x, x\rangle - \langle Kx,x\rangle = a ||x||^2 - ||x||^2_1$. Thus require: $$a \geq \frac{||x||^2_1}{||x||^2},\ \forall x \in \mathscr{H}\backslash\{0\}.$$ So we need to prove the supremum of the RHS of the above expression is equal to $||K||$. From Cauchy-Schwarz inequality, we have: $$\frac{||x||^2_1}{||x||^2} = \frac{\langle Kx, x\rangle}{||x||^2} \leq \frac{||K||\cdot||x||^2}{||x||^2} = ||K||. $$ However I am struggling to prove the reverse inequality.
Edit:
The definition my textbook is using does not require an inner product $\langle \cdot \, \ \cdot \rangle$ to satisfy the condition $\langle x, x \rangle = 0$ if and only if $x = 0$. When this condition is satisfied, it is denoted as a definite inner product instead.