$||x||^2/||A^{-1}||\leq x^TAx$ for positive definite symmetric matrix $A$

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I am trying to prove the inequality on Page 605 of Nocedal and Wright (Numerical Optimization);

$$\sigma_n(A)||x||^2 = ||x||^2/||A^{-1}||\leq x^TAx \leq ||A|| ||x||^2 = \sigma_1(A) ||x||^2$$

where $A$ is a symmetric positive definite matrix and $\sigma_n(A)$ is the $nth$ singular value of $A$ and $\sigma_1(A)$ is the first singular value of $A$.

I have been able to prove every part of the inequality except for;

$$||x||^2/||A^{-1}||\leq x^TAx$$

If someone could provide a proof or a reference to a book that proves this it would be greatly appreciated.

Below are the details of the proofs of the other parts of the inequality;

First here is the proof of $x^TAx\leq ||x||^2 \cdot ||A||$;

$$\begin{aligned} x^TAx &= ||x|| \cdot ||Ax|| \cos (\theta) \\ &= ||x||^2 \cdot \frac{||Ax||}{||x||} \cos(\theta)\\ &\leq ||x||^2 \max_{x \neq 0} \frac{||Ax||}{||x||} \quad \quad \cos(\theta)\geq 0 \text{ since } x^TAx \geq 0\\ &= ||x||^2 \cdot ||A|| \end{aligned}$$

and also here is the proof of $$||x||^2/||A^{-1}||\leq ||A|| \cdot ||x||^2$$;

$$\begin{aligned}||x||^2/||A||^{-1} &=||x||^2 \min_{y \neq 0} \frac{||y||}{||A^{-1}y||} \\ &= ||x||^2 \min_{y \neq 0} \frac{||AA^{-1}y||}{||A^{-1}y||} \\ & \leq ||x||^2 \min_{y \neq 0} \frac{||A|| \cdot ||A^{-1}y||}{||A^{-1}y||} \\ &= ||x||^2 ||A|| \end{aligned}$$

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Since $A$ is symmetric positive definite, we can write $A = \sum_k \sigma_k v_k v_k^T$, and we can assume the $\sigma_k$ are ordered.

It is straightforward to check by multiplying that $A^{-1} = \sum_k {1 \over \sigma_k} v_k v_k^T$, and so $\|A^{-1}\| = {1 \over \sigma_n}$.

Then $x^TAx = \sum_k \sigma_k ( v_k^T x)^2$ and so $\sigma_n \sum_k ( v_k^T x)^2 \le x^T Ax \le \sigma_1 \sum_k ( v_k^T x)^2$.

Since the $v_k$ are orthonormal, $\sum_k ( v_k^T x)^2 = \|x\|^2$.