Does the inequality $\frac{p^Tp}{p^TAp} < 2\frac{p^TA^{-1}p}{p^Tp}$ hold?

Question

Consider $A \in \mathbb{R}^{m \times m}$, $b\in\mathbb{R}^m$, $p \in \mathbb{R}^m$ and $x\in\mathbb{R}^m / \{A^{-1}b\}$, were the following properties and relations hold:

1. $\quad A>0$
2. $\quad A^T=A$
3. $\quad p:=b-Ax$
4. $\quad \lambda_{max}(A)=\lambda_{min}^{-1}(A^{-1})$
5. $\quad \lambda_{min}(A)|x|^2\leq x^TAx\leq \lambda_{max}(A)|x|^2$
6. $\quad$ We may want to limit us to $m=2$ if this helps to prove the inequality.

I want to prove (or in the worst case disprove), that the inequality $$\frac{p^Tp}{p^TAp} < 2\frac{p^TA^{-1}p}{p^Tp}$$ hold. I will sketch my attempts, since they may be helpful for other approaches on the problem:

First attempt: $$2\cdot \frac{p^TA^{-1}p}{p^Tp} > \frac{p^TA^{-1}p}{p^Tp} \geq \lambda_{min}(A^{-1})\cdot\frac{|p|^2}{|p|^2}=\frac{1}{\lambda_{max}(A)}\cdot\frac{|p|^2}{|p|^2}\leq \frac{p^Tp}{p^TAp} $$ This attempt ends here, since the last estimation goes into the 'wrong' direction.

Second attempt: $$\frac{p^Tp}{p^TAp} < 2\frac{p^TA^{-1}p}{p^Tp} \\ \Leftrightarrow\quad p^Tp\cdot p^Tp<2\cdot p^TAp\cdot p^TA^{-1}p \\ \Leftrightarrow\quad |p|_{pp^T}^2<2\cdot |p|_{App^TA^{-1}}^2$$ Here, $|z|_C=z^TCz=|C^{1/2}z|, C>0, C^T=C$ denote a weighted vector norm. At this step i'm not sure how to continue. For once, the matrix $pp^T$ is of rank 1, $\lambda_{max}(pp^T)=p^Tp$ and $\lambda_{min}(pp^T)=0$, since $pp^T$ is singular. The matrix $App^TA^{-1}$ is of rank 1 as well, since the transformation with $A$ wont change the eigenvalues and thus $\lambda_{max}(App^TA^{-1})=p^Tp$ and $\lambda_{min}(App^TA^{-1})=0$. Maybe it's possible to work with those weighted vector norms, without using the lower / upper bound ($\lambda_{min}$ and $\lambda_{max}$).

Third attempt:

I tried to find any counterexamples in order to disprove the inequality, but a could not find one.

Answer 1

Since $A$ is positive-define symmetric, then there exist an orthonormal basis $e_1,\ldots,e_m$ and numbers $\lambda_1, \ldots, \lambda_m$ such that $A e_i = \lambda_i e_i$. Denote $p_i = \langle p,e_i \rangle$, we have $ p= p_1 e_1 + \cdots + p_m e_m$, $p^T p = p_1^2 + \cdots + p_m^2$, $p^T A p = \lambda_1 p_1^2 + \cdots + \lambda_m p_m^2$ and $p^T A^{-1} p = \lambda_1^{-1} p_1^2 + \cdots + \lambda_m^{-1} p_m^2$. Hence, by Cauchy-Schwartz inequality, we have $$p^T Ap\,\times\, p^T A^{-1}p =(\sum_{i=1}^m \lambda_i p_i^2) (\sum_{i=1}^m \lambda_i^{-1} p_i^2) \geq (\sum_{i=1}^m p_i^2)^2 = (p^Tp)^2.$$