# What is $\max\limits_{\sum_{i=k}^{n}x_i\leq\sum_{i=k}^{n}y_i \\\forall k=1,2,\cdots,n} \prod_{i=1}^{n} x_i$

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

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 $$x_i=y_i$$?

• 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 $k=n$ – Lee Nov 19 '18 at 8:13