$(x^TAx)(x^TA^{-1}x)\leq \max_{1\leq i \leq n} \left( c\lambda_i +\frac{1}{c\lambda_i} \right)$ for any $c >0$ if $A$ is s.p.d. Suppose that $A$ is an $n \times n$ symmetric positive definite matrix. Assume that its eigenvalues are $\lambda_1 \leq \ldots \leq \lambda_n$. I would like to show that
$$(x^TAx)(x^TA^{-1}x)\leq \max_{1\leq i \leq n} \left(c\lambda_i +\frac{1}{c\lambda_i}    \right)$$ 
for any $c >0$.
This inequality seems quite strong since it is true for any $c>0$. If there is an error, please let me know.
Since $A$ is symmetric, by the Real Spectral Theorem, I know that there exists a basis of $\mathbb{R}^n$ consisting of orthonormal eigenvectors $e_1, \ldots, e_n$ of $A$. Suppose that $\lambda_1 \leq \ldots \leq \lambda_n$ are the corresponding eigenvalues of $e_1, \ldots, e_n$, respectively. 
I know that
$$(x^TAx)(x^TA^{-1}x)= \left( \sum_{i=1}^n (e_i^Tx)^2\lambda_i  \right)  \left( \sum_{i=1}^n (e_i^Tx)^2\frac{1}{\lambda_i}  \right).$$
Note that $\lambda_i>0$ for each $1 \leq i \leq n$ since $A$ is positive definite. 
This is where I get stuck. I have been told that I can use the Cauchy Schwarz Inequality and the arithmetic geometric mean inequality to obtain the result, but I keep reaching a dead end. 
Any help is extremely appreciated, thank you.
 A: For ease of notation I will handle the case where $A=\text{diag}(\lambda_1, \ldots, \lambda_n)$. The argument (once completed...) can easily be modified to the general case by a change of basis.
\begin{align}
x^\top A x x^\top A^{-1} x
&= x^\top (cA) x x^\top (cA)^{-1} x
\\
&= \left(\sum_i c\lambda_i x_i^2\right)\left(\sum_i \frac{1}{c\lambda_i} x_i^2\right)
\\
&\le \frac{1}{4} \left(\sum_i \left(c\lambda_i + \frac{1}{c\lambda_i}\right) x_i^2\right)^2 & \text{AM-GM}
\\
&\le \frac{1}{4} \left(\sum_i x_i^2\right)^2 \max_i \left(c\lambda_i + \frac{1}{c\lambda_i}\right)^2
\\
&= \frac{1}{4} \max_i \left(c\lambda_i + \frac{1}{c\lambda_i}\right)^2 & \text{if $\|x\|_2 \le 1$}
\end{align}
My final answer looks slightly different than yours, but perhaps I have made some errors...

Subsequent details, in response to comments below. (Details sketched out by A.Γ., thanks!)
Because $f(z) := z + \frac{1}{z}$ is a convex function, we have
$$g(c) := \max_i \left(c \lambda_i + \frac{1}{c \lambda_i}\right) = \max_i f(c\lambda_i) = \max\{f(c\lambda_1), f(c \lambda_n)\}.$$
To get the best upper bound, we want to minimize $g$ with respect to $c$. 
As claimed below, the minimizer, $c_*$, is the solution to $f(c \lambda_1) = f(c\lambda_n)$.
Then we obtain $c_*=(\lambda_1 \lambda_n)^{-1/2}$ and the final value is $\sqrt{\lambda_1/\lambda_n} + \sqrt{\lambda_n/\lambda_1}$ which yields the Kantorovich inequality.
To see why solving $f(c \lambda_1) = f(c\lambda_n)$ yields the minimizer, you can note that both $c \mapsto f(c\lambda_1)$ and $c \mapsto f(c\lambda_n)$ are convex functions that have the same minimum value $2$ at minimizers $c=1/\lambda_1$ and $c=1/\lambda_n$ respectively. You can draw some pictures to get the intuition for why solving $f(c \lambda_1) = f(c\lambda_n)$ yields the minimizer and then make this rigorous.
