Find the largest c>0 such that for all n \geq 1 and a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n}>0 we have
\sum_{j=1}^{n} a_{j}^{4} \geq c \sum_{k=1}^{n} \frac{\left(\sum_{j=1}^{k} a_{j} b_{k+1-j}\right)^{4}}{\left(\sum_{j=1}^{k} b_{j}^{2} j!\right)^{2}}