How to prove $\prod_{i=1}^{r}\left(1+\frac{1}{x_{i}}\right)\le \frac{2^{2^r}-1}{2^{2^r-1}}$? 
Let $x_{1},x_{2},\cdots,x_{r}$ be positive integers  such that
  $$1\le x_{1}\le x_{2}\le \cdots\le x_{r}$$
  and
  $$\prod_{i=1}^{r}\left(1+\dfrac{1}{x_{i}}\right)<2.$$
then Show  that
  $$\prod_{i=1}^{r}\left(1+\dfrac{1}{x_{i}}\right)\le \dfrac{2^{2^r}-1}{2^{2^r-1}}\tag{1}$$
  if and only if $x_{i}=2^{2^{i-1}}$

I know about the following similar problem (called the Erdos conjecture,and I know it has been solved)
if  the $a_n$ are postive integers such that
$$\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}<1$$
then find the maximum value of
$$
\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}$$
and the solution to this problem is given by the sequence $\{r_{n}\}$ defined by
$$r_{1}=2,r_{n}=r_{1}r_{2}\cdots r_{n-1}+1.n\ge 2$$
we have
$$\max{\left(\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}\right)}\le\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}+\cdots+\dfrac{1}{r_{n}}$$
the full solution can see  China Team Selection Test 1987 :http://www.artofproblemsolving.com/Forum/viewtopic.php?p=234089&sid=f7910d60017435727b0f23367e47f02a#p234089
Thank you 
For inequality $(1)$, I tried using the recursion $r_{n}=(r_{n-1})^2,\,r_{1}=2$, but failed.Thank you 
 A: I think, I'm able to show equality in equation $(1)$ if $x_i = 2^{2^{i-1}}$. This immediately shows one way of the iff statement. 
First note, that 
\begin{align}
\sum_{i=0}^{r-1} 2^i=\sum_{i=1}^{r} 2^{i-1}= 2^r-1
\end{align}
which I consider given ( maybe proof via induction ).
When writing $\log$, I will always refer to $\log_2$ ( since $ld$ is no Tex command)
Now, assume $x_i$ is defined as above.
Then taking the logarithm of your product and applying the logarithm laws, we get
\begin{align}
\log \prod_{i=1}^r (1+1/x_i) &= \sum_{i=1}^r \log (1+1/x_i) = \sum_{i=1}^r \log ((x_i+1)/x_i) 
\\
&=\sum_{i=1}^r \log (x_i+1)-\sum_{i=1}^r \log (x_i) \\
&=\sum_{i=1}^r \log (2^{2^{i-1}}+1)-\underbrace{\sum_{i=1}^r \log (2^{2^{i-1}}) }_{} \\
&=\sum_{i=1}^r \log (2^{2^{i-1}}+1)-\underbrace{\sum_{i=1}^r {2^{i-1}} }_{}\\
&=\underbrace{\sum_{i=1}^r \log (2^{2^{i-1}}+1)}_{}-(2^{r}-1) \\
&=\log\prod_{i=1}^r(2^{2^{i-1}}+1)-(2^{r}-1) \\
&=\log A_p-(2^{r}-1) \\
\end{align}
with $A_p=\prod_{i=1}^r(2^{2^{i-1}}+1)$. By expanding $A_p$ and summing all possible combinations ( one might want to proof this in detail ) we get 
\begin{align}
A_p=\prod_{i=1}^r(2^{2^{i-1}}+1) = \sum_{k=1}^{2^r}2^{k-1}
\end{align}
with the "Note" from above we then have $A_p = 2^{2^{r}}-1$. 
We then obtain
\begin{align}
&\log \prod_{i=1}^r (1+1/x_i)  = \log(2^{2^{r}}-1)-(2^{r}-1) \\
\Leftrightarrow& \prod_{i=1}^r (1+1/x_i) = 2^{{log(2^{2^{r}}-1)}-{(2^{r}-1) }}
= \frac{2^{2^{r}}-1}{2^{2^{r}-1}}
\end{align}
So, obviously we have:


*

*if $x_i = 2^{2^{i-1}}$, then $\prod_{i=1}^r (1+1/x_i)\leq \frac{2^{2^{r}}-1}{2^{2^{r}-1}}$ (actually we have equality)

*if $\prod_{i=1}^r (1+1/x_i)= \frac{2^{2^{r}}-1}{2^{2^{r}-1}}$, then $x_i = 2^{2^{i-1}}$


the problem is the second dot, since we actually had to show


*

*if $\prod_{i=1}^r (1+1/x_i)\leq \frac{2^{2^{r}}-1}{2^{2^{r}-1}}$, then $x_i = 2^{2^{i-1}}$


Here I don't really know what to do. But giving the fact that we have equality, you might be able to come up with something I wasn't able to derive.
Good luck!
A: Take $r = 1$ and $x_1 = 3$. Then, certainly, $(1 + 1/x_1) = 4/3 < 2$ but 
\begin{equation}
\frac{2^2 - 1}{2^{2 - 1}} = \frac{3}{2} \geq 4/3
\end{equation}
and $3$ is certainly not a power of $2$ so your statement is false, surely? I might be missing something/being incredibly stupid.
Moreover, the expression on the right is independent of the actual choice of $x_i$ but, if we increase the $x_i$, the product decreases so, if we find a sequence $x_1, x_2, \dots, x_n$ that works then, surely, $x_1 + 1, x_2 + 1, \dots x_n + 1$ also works.
