Let $u,v$ be the unit vectors and $A\in\mathscr{B(\mathscr{H})}$. Is the following inequality true $$\vert\langle Au,v\rangle\vert\leq \frac{w(A)}{\vert\langle u,v\rangle\vert}$$ where $w(A):=\sup\limits_{\Vert f\Vert=1}\vert\langle Af,f\rangle\vert$ is the numerical radius of A
Comments: First note that if $0\leq\vert\langle u,v\rangle\vert\leq\frac{1}{2}$, the inequality follows as $\Vert A\Vert\leq 2w(A)$. The inequality also follows if $\vert\langle u,v\rangle\vert=1$. But I could neither to prove the inequality for $\frac{1}{2}<\vert\langle u,v\rangle\vert<1$ nor find a counter example to disprove it.
Any comment/hint is highly appreciated. Thanks in advance.
The inequality does not hold in general. Let $H=\mathbb C^2$, $$ A=\begin{bmatrix}0&1\\0&0\end{bmatrix},\qquad u=\begin{bmatrix}1/2\\ \sqrt3/2\end{bmatrix},\qquad v=\begin{bmatrix} \sqrt3/2\\1/2\end{bmatrix}. $$ Then $\omega(A)=1/2$, $$ \langle Au,v\rangle=\frac{\sqrt3}2\frac{\sqrt3}2=\frac34, $$ $$ \langle u,v\rangle=\frac{\sqrt3}2, $$ and $$ \frac{\omega(A)}{\langle u,v\rangle}=\frac{1/2}{\sqrt3/2}=\frac1{\sqrt3}<\frac34=\langle Au,v\rangle. $$