The rank of a matrix: $A+A^t=J-I$

Question

Let $A=(a_{ij})$ be a $n\times n$ real matrix that

$$a_{ij}+a_{ji}=\begin{cases} 1& i\neq j\\ 0& i=j \end{cases}$$

Prove: $\text{rank}(A)\geq n-1$.

My attempts

The restriction on $A$ is equivalent to $A+A^t=J-I$ where $J$ is the matrix whose entries are $1$ and $I$ is the identity matrix.

1. Let $x,y$ be $n\times 1$ nonzero vectors such that $Ax=Ay=0$. I need $x=ky$. Since $Ax=Ay=0$, we have $$0=x^t(A+A^t)y=x^t(J-I)y=\sum_{i=1}^n x_i\sum_{i=1}^n y_i-\sum_{i=1}^n x_i y_i.$$ This is the first time that I stuck.

2. I try to use induction on $n$ to fine at least $n-1$ independent row vectors of $A$. The IH tells us that the first $n-1$ rows will contribute at least $n-2$ such vectors. However, there are two possibilities for the last one: the last column making the first $n-1$ rows independent or the vector on the last row. IDK how to start the calculation.

Thanks!