I would like to solve the below optimization problem. Any hints is appreciated. I'm gussing the answer is 1. \begin{equation} \begin{aligned} & \min \prod_{i=1}^{i=3}\left(\frac{1+x+ay}{1+x_i+ay_i}\right)^{p_i} \\ & \operatorname{s.t.} & & \sum_{i=1}^{3}p_i =1, \\ &&& \sum_{i=1}^{3} p_ix_i \le x, \\ &&& \sum_{i=1}^{3} p_iy_i \le y,\\ &&&p_i\ge 0, \; x_i\ge 0, \;y_i\ge 0,\; x> 0, \;y >0. \end{aligned} \end{equation} in which $p_i, x_i$, and $y_i$ are variables and $a$, $x$, and $y$ are constant.
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- $\begingroup$ $ x $ and $ y $ are greater than 0. $\endgroup$Bob– Bob2015-09-21 19:29:44 +00:00Commented Sep 21, 2015 at 19:29
- $\begingroup$ Well, The quotient needs to be min., so the Denominator needs to be max. so $x_i+ay_i$ needs to be max. The answer of 1 sounds true, but I guess you need a numerical algorithm for this one. $\endgroup$NoChance– NoChance2015-09-21 22:42:14 +00:00Commented Sep 21, 2015 at 22:42
- $\begingroup$ You need to tell us what are the variables and what are the constants. $\endgroup$Jerry Guern– Jerry Guern2015-09-21 23:22:19 +00:00Commented Sep 21, 2015 at 23:22
- $\begingroup$ $p_i, x_i$, and $y_i$ are variables and $a$, $x$, and $y$ are constant. $\endgroup$Bob– Bob2015-09-21 23:42:54 +00:00Commented Sep 21, 2015 at 23:42
- 1$\begingroup$ Numerical test indeed indicates that the optimal solution is $x_i = x, y_i = y, p_i = 1/3$. $\endgroup$Johan Löfberg– Johan Löfberg2015-09-22 06:41:34 +00:00Commented Sep 22, 2015 at 6:41
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1 Answer
$\begingroup$ $\endgroup$
We find the minimum of a more general expression
$$E=\prod_{i=1}^{n}\left(\frac{1+x+ay}{1+x_i+ay_i}\right)^{p_i},$$ where we substituted the number $3$ by for any natural number $n$. If some of our $p_i$’s are zero, we can drop them and diminish $n$. Also we recall Weighted AM-GM inequality: Let $p_1,\dots, p_n$ and $t_1,\dots, t_n$ be positive number such that $\sum_{i=1}^n p_i=1$. Then
$$\sum_{i=1}^n p_it_i\ge \prod_{i=1}^n t_i^{p_i}.$$
Now we remark that the problem to find a minimum of $E$ is equivalent to the problem to to find a maximum of $E^{-1}$. But
$$E^{-1}=\prod_{i=1}^{n}\left(\frac{1+x_i+ay_i}{1+x+ay}\right)^{p_i}=$$ $$\prod_{i=1}^{n}\left(1+x_i+ay_i\right)^{p_i}\left(\prod_{i=1}^{n}\left(1+x+ay\right)^{p_i}\right)^{-1}=$$ $$\prod_{i=1}^{n}\left(1+x_i+ay_i\right)^{p_i}\left(1+x+ay\right)^{-\sum_{i=1}^n p_i}=$$ $$\prod_{i=1}^{n}\left(1+x_i+ay_i\right)^{p_i}\left(1+x+ay\right)^{-1}.$$
But by Weighted AM-GM inequality $$\prod_{i=1}^{n}\left(1+x_i+ay_i\right)^{p_i}\le \sum_{i=1}^n p_i(1+x_i+ay_i)\le 1+x+ay.$$
So $E^{-1}\le 1$ and the equality is attained, for instance when all $x_i$’s are equal to $x$, all $y_i$’s are equal to $y$, and all $p_i$’s are equal to $1/n$.
