Find "almost inverse" of positive definite bilinear form Let $A$ be a positive definite $d \times d$ matrix, and define $A(x,x)=x^TAx$. Let $x$ be a point such that $\vert x^T\xi\vert^2\leq \xi^T A\xi$ for all $\xi\in\mathbb{R}^d$. Is this somehow equivalent to 
$A^{-1}(x,x)=x^TA^{-1}x\leq 1$?
 A: Yes, it's true. We may assume that $A=diag(\lambda_1,\cdots,\lambda_n)$ where $\lambda_j>0$ and let $X=[x_1,\cdots,x_n]^T$. The condition is: for every $\xi$, $\sum_i\lambda_i\xi_i^2-(\sum_ix_i\xi_i)^2\geq 0$ or $\sum_i(\lambda_i-x_i^2)\xi_i^2-2\sum_{i<j}x_ix_j\xi_i\xi_j\geq 0$, that is equivalent to $B\geq 0$ where $B=diag(\lambda_1,\cdots,\lambda_n)-XX^T=A-XX^T$. In particular $\det(B)\geq 0$. According to Sherman-Morrison, $\det(B)=(1-X^TA^{-1}X)\det(A)\geq 0$, that is equivalent to $X^TA^{-1}X\leq 1$ and we are done.
EDIT. The converse is also true. Proof. Note that $XX^T$ has only one non-zero eigenvalue $X^TX=||X||^2$ which is associated to the eigenvector $X$. Then $A\geq XX^T$ iff $X^TAX\geq ||X||^4$, that is $\sum_i\lambda_ix_i^2\geq ||X||^4$. We want to prove that: $\sum_i\dfrac{x_i^2}{\lambda_i}\leq 1$ implies that $\sum_i\lambda_ix_i^2\geq ||X||^4$.
We assume that $X$ is fixed and the $(\lambda_i)$ vary in $({\mathbb{R}^+}^*)^n$ and we seek $\inf \sum_i\lambda_ix_i^2$ under the condition $\sum_i\dfrac{x_i^2}{\lambda_i}= \sigma$ where $0\leq \sigma\leq 1$. We obtain easily that the $(\lambda_i)$ are equal and that the lower bound is reached for $\sigma=1$ and is $||X||^4$; then we are done.
