How to prove $\prod_{i=1}^{r}\left(1+\frac{1}{x_{i}}\right)\le \frac{2^{2^r}-1}{2^{2^r-1}}$?

Question

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

Answer 1

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!

Answer 2

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.