$\frac{\prod_{i=1}^n (1+x_i)-1}{\prod_{i=1}^n (1+x_i/\delta)-1} \stackrel{\text{?}}{\le} \frac{(1+x_n)^n-1}{(1+x_n/\delta)^n-1} $ . Let $x_1 \le x_2 \le \cdots \le x_n$. Let $\delta>1$ be some positive real numbers. I assume that $0\le x_i <1$, for $i=1,\ldots,n$ and $x_n >0$.
Does the following expression hold?
$$ \frac{\prod_{i=1}^n (1+x_i)-1}{\prod_{i=1}^n (1+x_i/\delta)-1} \stackrel{\text{?}}{\le} \frac{(1+x_n)^n-1}{(1+x_n/\delta)^n-1} $$

By putting:
$$A=\sum_{i=1}^n\log(1+x_i),\quad A_\delta=\sum_{i=1}^n\log(1+x_i/\delta),$$
$$B=\sum_{i=1}^n\log(1+x_n),\quad B_\delta=\sum_{i=1}^n\log(1+x_n/\delta),$$
the inequality is equivalent to:
$$e^{A+B_\delta}+e^{A_\delta}+e^{B}\leq e^{B+A_\delta}+e^{A}+e^{B_\delta}.$$
Provided that $A+B_\delta\leq B+A_\delta$, the latter follows from Karamata's inequality.
So we have only to show that:
$$ B-A \geq B_\delta-A_\delta, \tag{1}$$
i.e.
$$ \sum_{i=1}^n\log\left(\frac{1+x_n}{1+x_i}\right)\geq \sum_{i=1}^n\log\left(\frac{\delta+x_n}{\delta+x_i}\right).\tag{2}$$
This is quite trivial since for any $1>C>D>0$ the function $f(x)=\log\left(\frac{x+C}{x+D}\right)$ is positive and decreasing on $\mathbb{R}^+$ since its derivative is $\frac{D-C}{(x+C)(x+D)}<0$, $f(0)>0$ and $\lim_{x\to+\infty}f(x)=0.$
