Suppose $A$ is a $d \times d$ symmetric matrix, $A_{i,j}$ be a $(i,j)$ entry of the matrix $A$, and $\Vert \cdot \Vert _{op}$ be a matrix norm. Then, show that $$\Vert A \Vert _{op} \le \max_{1 \le i \le d} \sum_{j = 1}^d |A_{i,j}|$$
My goal is to prove the above statement.
I first use the definition of the operator norm as follows;
$$\Vert A \Vert _{op} = \max_{v:\Vert v\Vert_2 = 1}\Vert Av\Vert_2$$
$\Vert \cdot \Vert _p \le \Vert \cdot \Vert _q$ for $1 \le q \le p$,
so we have $\Vert A \Vert _{op} \le \max_{v:\Vert v\Vert_2 = 1}\Vert Av\Vert_1$
The inequality I have to prove has some expression associated with $l_1$ norm, so I tried to induce the above inequality.
This is where I'm stuck, and I'm not sure how to proceed from here (and whether this approach is correct).
Any idea regarding this proof would be grateful.