Let $A$ be a $n\times n$ complex matrix. Define the numerical norm of $A$ as $$w(A)=\sup\{|x^*Ax|;\|x\|_2=1\}, \|x\|_2^2=\sum_{i=1}^n|x_i|^2.$$ And the spectral norm of $A$ is $$\|A\|_\infty =\sup_{\|x\|_2=1}\|Ax\|_2.$$ Then we have $$\frac12\|A\|_\infty \leq w(A)\leq \|A\|_\infty.$$
I do know how to prove $w(A)\leq \|A\|_\infty$, which is easy. However, I could not prove the left inequality, in particular, the factor $1/2$ is hard to think...
Notice that the matrix need not to be symmetric.
Let $A=H+S$, where $H:=(A+A^*)/2$ and $S:=(A-A^*)/2$. The hermitian and skew-hermitian parts $H$ and $S$ of $A$ are normal matrices and for normal matrices, the spectral norm coincides with the spectral radius which is equal to the numerical radius. Hence $$ \|A\|\leq\|H\|+\|S\|=w(H)+w(S). $$ We have $$ \begin{split} \|A\| &\leq w(H)+w(S)\\ &= \frac{1}{2}\left(\sup_{\|x\|=1}|x^*(A+A^*)x|+\sup_{\|x\|=1}|x^*(A-A^*)x|\right) \\ &\leq \frac{1}{2}\left(2\sup_{\|x\|=1}|x^*Ax|+2\sup_{\|x\|=1}|x^*Ax|\right)=2w(A). \end{split} $$
