To prove Kantorovich inequality (for that we suppose the matrix A symmetric and definite positive) I need to demonstrate the next exercise:
Prove that
$$(x^TAx)(x^TA^{-1}x) \leq \frac{(\lambda_1 + \lambda_n)^2}{4\lambda_1 \lambda_n} || x||_2^4 \quad \forall x \in \mathbb{R}^n $$
is equivalent to
$$(x^TAx)(x^TA^{-1}x) \leq \frac{(\lambda_1 + \lambda_n)^2}{4\lambda_1 \lambda_n} \quad \forall x \in \mathbb{R}^n,||x||_2=1 $$
The part that requires work is $\Leftarrow$.
Suppose that $(x^TAx)(x^TA^{-1}x) \leq \frac{(\lambda_1 + \lambda_n)^2}{4\lambda_1 \lambda_n} \quad \forall x \in \mathbb{R}^n,||x||_2=1$
Let $x\in \mathbb R^n\setminus\{0\}$
Let $y=\frac{x}{||x||}$
Since $||y||=1$, one has $$(y^TAy)(y^TA^{-1}y)\leq \frac{(\lambda_1 + \lambda_n)^2}{4\lambda_1 \lambda_n}$$
which is equivalent to $$\frac{1}{||x||^4}(x^TAx)(x^TA^{-1}x) \leq \frac{(\lambda_1 + \lambda_n)^2}{4\lambda_1 \lambda_n}$$
This proves the claim.
Also, http://www.math.rice.edu/~njd2/documents/inequalities.pdf might improve your understanding of scaling in inequalities.
