Let $T$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$. It is known that $$ \sup_{\|x\|=1} \|Tx\| = \sup_{\|x\|=1} |(Tx,x)|, $$ see for instance the answer given here.
Is there a way to prove that the same equality holds with supremum replaced by infimum on both sides? Using Cauchy-Schwarz we easily obtain one inequality, so it remains to prove that
$$ \inf_{\|x\|=1} \|Tx\| \le \inf_{\|x\|=1} |(Tx,x)|. $$ The standard strategy in the sup case is to use polarization identity for $|(Tx,y)|$, but it seems that this now eventually leads to the converse inequality.
I would have liked to find something neater, but here is an argument.
Suppose first that $T$ is of the form $$\tag1 T=\sum_j\lambda_jP_j, $$ where $\{P_j\}$ are pairwise orthogonal projections that add to the identity. Then $$ \|Tx\|^2=\Big\|\sum_j\lambda_jP_jx\Big\|^2=\sum_j|\lambda_j|^2\,\|P_jx\|^2. $$ It follows (by taking $x$ in the range of $P_j$ corresponding to $j$ with small $|\lambda_j|$) that $$ \inf_{\|x\|=1}\|Tx\|=\inf\{|\lambda_j|:\ j\}. $$ Also, $$ \langle Tx,x\rangle=\sum_j\lambda_j\|P_jx\|^2. $$ So $$ \inf_{\|x\|=1}|\langle Tx,x\rangle|=\inf\{|\lambda_j|: j\}=\inf_{\|x\|=1}\|Tx\|. $$ For arbitrary $T$, fix $\varepsilon>0$. Then there exists $T'$, of the form $(1)$, with $\|T-T'\|<\varepsilon$. As $$ \big|\,\|Tx\|-\|T'x\|\,\big|\leq\|Tx-T'x\|\leq\|T-T'\|\,\|x|<\varepsilon\,\|x\| $$ and $$ |\langle Tx,x\rangle-\langle T'x,x\rangle|=|\langle (T-T')x,x\rangle|\leq\varepsilon \,\|x\|, $$ we get that $$ \inf_{\|x\|=1}|\langle Tx,x\rangle|<\varepsilon+\inf_{\|x\|=1}|\langle T'x,x\rangle|=\varepsilon+\inf_{\|x\|=1}\|T'x\|\leq2\varepsilon+\inf_{\|x\|=1}\|Tx\|. $$ Similarly, $$ \inf_{\|x\|=1}\|Tx\|\leq\varepsilon+\inf_{\|x\|=1}\|T'x\|=\varepsilon+\inf_{\|x\|=1}|\langle T'x,x\rangle|\leq2\varepsilon+\inf_{\|x\|=1}|\langle Tx,x\rangle|. $$ Thus $$ \big| \inf_{\|x\|=1}\|Tx\|-\inf_{\|x\|=1}|\langle Tx,x\rangle\big|\leq2\varepsilon. $$ As we are free to choose $\varepsilon$, this shows that $$ \inf_{\|x\|=1}\|Tx\|=\inf_{\|x\|=1}|\langle Tx,x\rangle|. $$