\begin{array}{ll} \text{maximize} & \prod_{i=1}^{n}x_i\\ \text{subject to} & \mathrm \sum_{i=k}^{n}x_i\leq\sum_{i=k}^{n}y_i \\\forall k=1,2,\cdots,n\end{array}

if $x_1\geq\cdots\geq x_n$ and $y_1\geq\cdots\geq y_n\quad$ ($x_i,y_i\in \mathbb{R}^+$ and all $y_i$ are given).

My attempt: By induction

For $n=2$, we need to maximize $x_1x_2$ when $x_1+x_2\leq y_1+y_2$ and $x_2\leq y_2$. Let $x_2=y_2-t\quad$ ($t\geq0$), then $x_1\leq y_1+t$, thus to maximize $x_1x_2$, we let $x_1=y_1+t$. Then $x_1x_2=(y_2-t)(y_1+t)=y_1y_2-(y_1-y_2)t-t^2\leq y_1y_2$. Thus $\max x_1x_2=y_1y_2$ when $t=0$.

Suppose $\max\limits_{{\sum_{i=k}^{n}x_i\leq\sum_{i=k}^{n}y_i \\\forall k=1,2,\cdots,n}} \prod_{i=1}^{n}x_i=\prod_{i=1}^{n}y_i$, then \begin{array}{ll}\max\limits_{{\sum_{i=k}^{n+1}x_i\leq\sum_{i=k}^{n+1}y_i \\\forall k=1,2,\cdots,n+1}} \prod_{i=1}^{n+1}x_i=(\prod_{i=1}^{n}y_i)x_{n+1}\leq(\prod_{i=1}^{n}y_i)y_{n+1}.\end{array}

I think I cannot use induction like that because of constraints, any other method to try to prove my hypothesis that maximum is achieved when $x_i=y_i$?

  • $\begingroup$ I cannot just let $k=1$, because inequality in the constraint should hold for all $k$, i.e. $x_1+\cdots+x_n\leq y_1+\cdots+y_n$ when $k=1$ till $x_n\leq y_n$ when $k=n$ $\endgroup$ – Lee Nov 19 '18 at 8:13

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