Well, at first glance recursive formula looks pretty ugly, so we think of telescoping. Define $S_i=a_0+a_1+...+a_i$ for $i=0,1,2...,n$. By plugging $k=0,1,...,n-1$ in formula and then adding we get: $$S_{n-1}=n \cdot c+\sum_{k=0}^{n-1} (\sum_{i=k}^{n-1} a_{i-k}(a_i+a_{i+1}) )$$From $a_n=0$ we have $S_n=\sum_{i=0}^n a_i=(\sum_{i=0}^{n-1}a_i)+a_n=(\sum_{i=0}^{n-1}a_i)=S_{n-1}$ so now $$S_{n}=n \cdot c+\sum_{k=0}^{n-1} (\sum_{i=k}^{n-1} a_{i-k}(a_i+a_{i+1}) )$$Now $\sum_{k=0}^{n-1} (\sum_{i=k}^{n-1} a_{i-k}(a_i+a_{i+1}))=\sum_{k=0}^{n-1} (a_0(a_k+a_{k+1})+a_1(a_{k+1}+a_{k+2})+...+a_{n-2-k}(a_{n-2}+a_{n-1})+a_{n-1-k}(a_{n-1}+a_n))$ $=a_0(a_0+a_1)+(a_0+a_1)(a_1+a_2)+(a_0+a_1+a_2)(a_2+a_3)+...=\sum_{i=0}^{n-1}((a_i+a_{i+1})\cdot (a_0+a_1+...+a_i) )=\sum_{i=0}^{n-1} ((S_{i+1}-S_{i-1}) \cdot S_i)$ Hence $\sum_{k=0}^{n-1} (\sum_{i=k}^{n-1} a_{i-k}(a_i+a_{i+1}))=\sum_{i=0}^{n-1} ((S_{i+1}-S_{i-1}) \cdot S_i)=S_1S_2-S_1S_2+S_2S_3-S_2S_3+...-S_{n-2}S_{n-1}+S_{n-2}S_{n-1}+S_{n-1}S_n=S_{n-1}S_n=S_n^2$. From previous we have:$$S_{n}=n \cdot c+\sum_{k=0}^{n-1} (\sum_{i=k}^{n-1} a_{i-k}(a_i+a_{i+1}) )=n \cdot c+S_n^2$$Then $-n \cdot c=S_n^2-S_n$ $\implies$ $-4n \cdot c+1=(2S_n-1)^2 \ge 0$ $\implies$ $4nc \le 1$ $\implies$ $c \leq \frac{1}{4n}$. Done. Equality holds when $S_n=\frac{1}{2}$