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What is $maxlimits_{sum_{i=k}^{n}x_ileqsum_{i=k}^{n}y_i \forall k=1,2,cdots,n} prod_{i=1}^{n} x_i$
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begin{array}{ll} text{maximize} & prod_{i=1}^{n}x_i\ text{subject to} & mathrm sum_{i=k}^{n}x_ileqsum_{i=k}^{n}y_i \forall k=1,2,cdots,nend{array}
if $x_1geqcdotsgeq x_n$ and $y_1geqcdotsgeq y_nquad$ ($x_i,y_iin mathbb{R}^+$ and all $y_i$ are given).
My attempt: By induction
For $n=2$, we need to maximize $x_1x_2$ when $x_1+x_2leq y_1+y_2$ and $x_2leq y_2$. Let $x_2=y_2-tquad$ ($tgeq0$), then $x_1leq y_1+t$, thus to maximize $x_1x_2$, we let $x_1=y_1+t$. Then $x_1x_2=(y_2-t)(y_1+t)=y_1y_2-(y_1-y_2)t-t^2leq y_1y_2$. Thus $max x_1x_2=y_1y_2$ when $t=0$.
Suppose $maxlimits_{{sum_{i=k}^{n}x_ileqsum_{i=k}^{n}y_i \forall k=1,2,cdots,n}} prod_{i=1}^{n}x_i=prod_{i=1}^{n}y_i$, then begin{array}{ll}maxlimits_{{sum_{i=k}^{n+1}x_ileqsum_{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 $x_i=y_i$?
optimization
asked 5 hours ago
Lee
706
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up vote
2
down vote
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begin{array}{ll} text{maximize} & prod_{i=1}^{n}x_i\ text{subject to} & mathrm sum_{i=k}^{n}x_ileqsum_{i=k}^{n}y_i \forall k=1,2,cdots,nend{array}
if $x_1geqcdotsgeq x_n$ and $y_1geqcdotsgeq y_nquad$ ($x_i,y_iin mathbb{R}^+$ and all $y_i$ are given).
My attempt: By induction
For $n=2$, we need to maximize $x_1x_2$ when $x_1+x_2leq y_1+y_2$ and $x_2leq y_2$. Let $x_2=y_2-tquad$ ($tgeq0$), then $x_1leq y_1+t$, thus to maximize $x_1x_2$, we let $x_1=y_1+t$. Then $x_1x_2=(y_2-t)(y_1+t)=y_1y_2-(y_1-y_2)t-t^2leq y_1y_2$. Thus $max x_1x_2=y_1y_2$ when $t=0$.
Suppose $maxlimits_{{sum_{i=k}^{n}x_ileqsum_{i=k}^{n}y_i \forall k=1,2,cdots,n}} prod_{i=1}^{n}x_i=prod_{i=1}^{n}y_i$, then begin{array}{ll}maxlimits_{{sum_{i=k}^{n+1}x_ileqsum_{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 $x_i=y_i$?
optimization
asked 5 hours ago
Lee
706
I cannot just let $k=1$, because inequality in the constraint should hold for all $k$, i.e. $x_1+cdots+x_nleq y_1+cdots+y_n$ when $k=1$ till $x_nleq y_n$ when $k=n$
– Lee
5 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
begin{array}{ll} text{maximize} & prod_{i=1}^{n}x_i\ text{subject to} & mathrm sum_{i=k}^{n}x_ileqsum_{i=k}^{n}y_i \forall k=1,2,cdots,nend{array}
if $x_1geqcdotsgeq x_n$ and $y_1geqcdotsgeq y_nquad$ ($x_i,y_iin mathbb{R}^+$ and all $y_i$ are given).
My attempt: By induction
For $n=2$, we need to maximize $x_1x_2$ when $x_1+x_2leq y_1+y_2$ and $x_2leq y_2$. Let $x_2=y_2-tquad$ ($tgeq0$), then $x_1leq y_1+t$, thus to maximize $x_1x_2$, we let $x_1=y_1+t$. Then $x_1x_2=(y_2-t)(y_1+t)=y_1y_2-(y_1-y_2)t-t^2leq y_1y_2$. Thus $max x_1x_2=y_1y_2$ when $t=0$.
Suppose $maxlimits_{{sum_{i=k}^{n}x_ileqsum_{i=k}^{n}y_i \forall k=1,2,cdots,n}} prod_{i=1}^{n}x_i=prod_{i=1}^{n}y_i$, then begin{array}{ll}maxlimits_{{sum_{i=k}^{n+1}x_ileqsum_{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 $x_i=y_i$?
optimization
asked 5 hours ago
Lee
706
begin{array}{ll} text{maximize} & prod_{i=1}^{n}x_i\ text{subject to} & mathrm sum_{i=k}^{n}x_ileqsum_{i=k}^{n}y_i \forall k=1,2,cdots,nend{array}
if $x_1geqcdotsgeq x_n$ and $y_1geqcdotsgeq y_nquad$ ($x_i,y_iin mathbb{R}^+$ and all $y_i$ are given).
My attempt: By induction
For $n=2$, we need to maximize $x_1x_2$ when $x_1+x_2leq y_1+y_2$ and $x_2leq y_2$. Let $x_2=y_2-tquad$ ($tgeq0$), then $x_1leq y_1+t$, thus to maximize $x_1x_2$, we let $x_1=y_1+t$. Then $x_1x_2=(y_2-t)(y_1+t)=y_1y_2-(y_1-y_2)t-t^2leq y_1y_2$. Thus $max x_1x_2=y_1y_2$ when $t=0$.
Suppose $maxlimits_{{sum_{i=k}^{n}x_ileqsum_{i=k}^{n}y_i \forall k=1,2,cdots,n}} prod_{i=1}^{n}x_i=prod_{i=1}^{n}y_i$, then begin{array}{ll}maxlimits_{{sum_{i=k}^{n+1}x_ileqsum_{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 $x_i=y_i$?
optimization
optimization
asked 5 hours ago
Lee
706
asked 5 hours ago
Lee
706
asked 5 hours ago
Lee
706
asked 5 hours ago
Lee
706
asked 5 hours ago
Lee
706
706
I cannot just let $k=1$, because inequality in the constraint should hold for all $k$, i.e. $x_1+cdots+x_nleq y_1+cdots+y_n$ when $k=1$ till $x_nleq y_n$ when $k=n$
– Lee
5 hours ago
add a comment |
I cannot just let $k=1$, because inequality in the constraint should hold for all $k$, i.e. $x_1+cdots+x_nleq y_1+cdots+y_n$ when $k=1$ till $x_nleq y_n$ when $k=n$
– Lee
5 hours ago
I cannot just let $k=1$, because inequality in the constraint should hold for all $k$, i.e. $x_1+cdots+x_nleq y_1+cdots+y_n$ when $k=1$ till $x_nleq y_n$ when $k=n$
– Lee
5 hours ago
I cannot just let $k=1$, because inequality in the constraint should hold for all $k$, i.e. $x_1+cdots+x_nleq y_1+cdots+y_n$ when $k=1$ till $x_nleq y_n$ when $k=n$
– Lee
5 hours ago
add a comment |
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I cannot just let $k=1$, because inequality in the constraint should hold for all $k$, i.e. $x_1+cdots+x_nleq y_1+cdots+y_n$ when $k=1$ till $x_nleq y_n$ when $k=n$
– Lee
5 hours ago