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{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. 
 A: 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 $1/n$.
