Let the partial sums $s_n$ of $\sum^{n}_{j=1} a_j$ be bounded above by $A > 0$ and the sequence $b_n$ satisfy $b_1\geq b_2\geq b_3...\geq b_n\geq 0$. I'm trying to show the following: $$ |\sum^{n}_{j = 1}a_jb_j| \leq 2Ab_1 $$ I tried breaking the summation up via "summation by parts" below and ended up with a different result than expected: $$\sum^{n}_{j=m+1}x_ jy_j=s_ny_{n+1}-s_my_m+\sum^{n}_{j=m+1}s_ j(y_j-y_{j+1})$$
$$|s_nb_{n+1}-s_mb_m+\sum^{n}_{j=1}s_ j(b_j-b_{j+1})| \leq |Ab_{n+1}-0*b_m+A\sum^{n}_{j=1}(b_j-b_{j+1})| \leq |Ab_{n+1}+Ab_1-Ab_{n+1})| \leq Ab_1$$ Which step did I mess up on? The one that seems most suspect was setting $s_0=0$ based on $x_1=s_1-s_0=s_1$ (to match lower bounds with the summation by parts formula), so if that's the issue I'd especially appreciate an explanation for why that doesn't work. Thanks!