# Any hints on how to solve this optimization problem?

I would like to solve the below optimization problem. Any hints is appreciated. I'm gussing the answer is 1. \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} in which $p_i, x_i$, and $y_i$ are variables and $a$, $x$, and $y$ are constant.

• $x$ and $y$ are greater than 0.
– Bob
Sep 21, 2015 at 19:29
• 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. Sep 21, 2015 at 22:42
• You need to tell us what are the variables and what are the constants. Sep 21, 2015 at 23:22
• $p_i, x_i$, and $y_i$ are variables and $a$, $x$, and $y$ are constant.
– Bob
Sep 21, 2015 at 23:42
• Numerical test indeed indicates that the optimal solution is $x_i = x, y_i = y, p_i = 1/3$. Sep 22, 2015 at 6:41

## 1 Answer

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 $1/n$.