Let $n \geq 2$ be an integer and $x_1, x_2, \cdots, x_n$ positive reals such that $x_1x_2 \cdots x_n = 1$. Show: $$\{x_1\} + \{x_2\} + \cdots + \{x_n\} < \frac{2n-1}{2}$$ Note: $\{x\}$ denotes the fractional part of $x$.

Is $\dfrac{2n-1}{2}$ optimal?

  • $\begingroup$ Would you provide the source? And of course, your own efforts, please. Thank you! $\endgroup$ – A. Pongrácz Aug 16 '18 at 10:04

Assume that $n\geq2$, $x_1x_2\cdots x_n=1$, and $$\{x_1\}+\{x_2\}+\ldots+\{x_n\}\geq n-{1\over2}\ .\tag{1}$$ Put $\tau_i:=1-\{x_i\}>0$. Then $(1)$ implies $\sum_i\tau_i\leq{1\over2}$; in particular no $\tau_i$ is $=1$. We therefore may write $x_i=m_i-\tau_i$ with $m_i=\lceil x_i\rceil\geq1$. As $\prod_i x_i=1$ at least one $m_i$ is $\geq2$. Now $$1=\prod_i x_i=\prod_i m_i\cdot\prod_i\left(1-{\tau_i\over m_i}\right)\ .\tag{2}$$ We shall need the following Lemma, which is easily proven using induction: If $n\geq2$, $x_i>0$ $(1\leq i\leq n)$, and $\sum_i x_i<1$ then $$\prod_{i=1}^n(1-x_i)>1-\sum_{i=1}^n x_i\ .$$Since $\sum_i{\tau_i\over m_i}<{1\over2}$ we then obtain from $(2)$ the contradiction $$1\geq 2\left(1-\sum_i{\tau_i\over m_i}\right)>2\cdot{1\over2}=1\ .$$


The bound is certainly optimal (if true): let $x_1= \frac{1}{2}+\varepsilon, x_n= 2-\varepsilon$, and all other $x_i= 1-\varepsilon$. (Not with the same number $\varepsilon$, just pick very small positive reals like this.)

This example also suggests the proof strategy. We may assume that none of the numbers are integers, as then a small perturbation leads to a significant improvement. Partition the indices so that $x_1, x_2, \ldots, x_k< 1$ and the rest is greater than $1$. This is possible with reindexing the elements if necessary. Note that because of the condition, $1\leq k\leq n-1$.

Assume that in the latter product, there are more than one terms that are greater than $2$. Then we can improve the objective function again: replace one such $x_i$ by $x_i-1$, and multiply all the numbers $x_1, x_2, \ldots, x_k$ by $\sqrt[k]{\frac{x_i}{x_i-1}}$. We have that the first $k$ numbers are still eactly the ones below $1$. By successive application of this step, we may assume that only one of the numbers that are greater than $1$ is greater than $2$.

A similar simplification can be made if $k\geq 2$. In that case, we can first replace $x_2$ by $1+\varepsilon$ and $x_1$ by $x_1x_2/(1+\varepsilon)$, iproving on the objective function.

So we may assume that $x_1<1$, $1<x_2, \ldots, x_{n-1}<2$ and $x_n>2$.

Further hints: try to push those middle elements close to $1$ while improving on the objective function.

Once you are done with that, the problem has essentially one varaible, as $x_1x_n\approx 1$, and you can get rid of the fractional parts easily in the objective function, to make it a simple case of finding the maximum of a univariate function.


Proof for $n=2$.

One and only one between $\lfloor{x_1}\rfloor$ and $\lfloor{x_2}\rfloor$ should be equal to $0$ since $x_1x_2=1$ and if both $\{x_1\}$ and $\{x_2\}$ are nul the proof follows immediately.

Put $x_1=k+\{x_1\}$ where $k\in\mathbb N^+$ and $\{x_1\}\gt0$ so $x_2=\{x_2\}=\dfrac{1}{k+\{x_1\}}$.

We have $\{x_1\}+\{x_2\}=\{x_1\}+\dfrac{1}{k+\{x_1\}}\lt\dfrac{2\cdot2-1}{2}=1+\dfrac12\space\color{red}{\large?}$

If $k\ge2$ then $\{x_1\}+\dfrac{1}{k+\{x_1\}}\le\{x_1\}+\dfrac 12$ so $\color{red}{\text{ YES}}$.

It remains to prove for $k=1$. In this case if $\{x_1\}+\dfrac{1}{1+\{x_1\}}=\dfrac32$ the positive number $\{x_1\}$ is the positive root of $$2X^2-X-1=0\Rightarrow\{x_1\}=1,\color{red}{ \text {absurde}}$$ A fortiori for $\{x_1\}+\dfrac{1}{1+\{x_1\}}\gt\dfrac32$ which end the proof.

Can someone from this get the general answer for $n\gt2$?


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