Understanding Kantorovich's inequality

I'm looking for a proof of the Kantorovich inequality, namely:

$$\langle Ax,x\rangle \langle A^{-1}x,x\rangle \leq \frac{1}{4}\left(K(A)+\frac{1}{K(A)}\right)$$ Where $K(A)= \lVert A\rVert_2 \lVert A^{-1}\rVert_2$ and $A$ is an Hermitian positive definite matrix and $x$ a vector with the accurate size. Or alternatively

$$\langle Ax,x\rangle \langle A^{-1}x,x\rangle \leq \frac{1}{4}\left(\left(\frac{\beta}{\alpha}\right)^{2}+\left(\frac{\alpha}{\beta}\right)^{2}\right)$$ where $0 < \alpha = \lambda_1 \leq \cdots \leq \lambda_n = \beta$ are the eigenvalues of $A$.

There are a lot of proofs on internet but this is the one that I found easier to understand. Nevertheless, I'm stuck figuring out something:

They use $f,g: [\alpha,\beta]\to \mathbb{R}$ two convex function with $f$ positive and $f(t)\leq g(t)$ for every $t \in [\alpha, \beta]$. Then they claim that $F=f(A)$ and $G=g(A)$ are well defined and hermitian positive definite.

I don't even understand why they mean by $f(A)$.

If any of you guys can suggest me alternative documentation (other proof) or help me with this one, I would be very grateful

• $F = f(A)$ is the matrix that is obtained by expandind $f(t)$ as a Taylor series (an "infinite polynomial") and then substituting $A$ for $t$ in that expansion. – NickD Apr 4 '18 at 20:13
• @Nick thank you but how does that implies that $F$ and $G$ are hermitian positive definite – Natalio Apr 4 '18 at 20:16
• If $\lambda$ is an eigenvalue of $A$ then $f(\lambda)$ is an eigenvalue of $f(A)$. I think that's enough to push through the proof, given the conditions. – NickD Apr 4 '18 at 20:39
• @Nick, yes that is enough. Thank you very much – Natalio Apr 4 '18 at 20:46

If $A$ is diagonalizable, i.e., if $A=SDS^{-1}$, with $D= \text{diag}(\lambda_1,\dots,\lambda_n)$, and if $f:\mathbb{R} \longrightarrow \mathbb{R}$, then $$f(A) := S f(D) S^{-1},$$ in which $f(D):= \text{diag}(f(\lambda_1),\dots,f(\lambda_n))$.
In the problem above, since every Hermitian matrix is diagonalizable (in fact, if $A$ is Hermitian, then there is a unitary matrix $U$ such that $U^*AU$ is diagonal), it follows that $f(A)$ is well-defined and positive definite since $f$ is positive.