Vtyjkyuk

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: $$x_1 + x_2 + cdots + x_n < frac2n-12$$
Note: $x$ denotes the fractional part of $x$.

Is $dfrac2n-12$ optimal?

Assume that $ngeq2$, $x_1x_2cdots x_n=1$, and
$$x_1+x_2+ldots+x_ngeq n-1over2 .tag1$$
Put $tau_i:=1-x_i>0$. Then $(1)$ implies $sum_itau_ileq1over2$; in particular no $tau_i$ is $=1$. We therefore may write $x_i=m_i-tau_i$ with $m_i=lceil x_irceilgeq1$. As $prod_i x_i=1$ at least one $m_i$ is $geq2$. Now
$$1=prod_i x_i=prod_i m_icdotprod_ileft(1-tau_iover m_iright) .tag2$$
We shall need the following Lemma, which is easily proven using induction: If $ngeq2$, $x_i>0$ $(1leq ileq n)$, and $sum_i x_i<1$ then
$$prod_i=1^n(1-x_i)>1-sum_i=1^n x_i .$$Since $sum_itau_iover m_i<1over2$ we then obtain from $(2)$ the contradiction
$$1geq 2left(1-sum_itau_iover m_iright)>2cdot1over2=1 .$$

The bound is certainly optimal (if true): let $x_1= frac12+varepsilon, x_n= 2-varepsilon$, 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, $1leq kleq n-1$.

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 $sqrt[k]fracx_ix_i-1$.
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 $kgeq 2$. 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 $x_1x_napprox 1$, 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 $lfloorx_1rfloor$ and $lfloorx_2rfloor$ 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 $kinmathbb N^+$ and $x_1gt0$ so $x_2=x_2=dfrac1k+x_1$.

We have $x_1+x_2=x_1+dfrac1k+x_1ltdfrac2cdot2-12=1+dfrac12spacecolorredlarge?$

If $kge2$ then $x_1+dfrac1k+x_1lex_1+dfrac 12$ so $colorredtext YES$.

It remains to prove for $k=1$. In this case if $x_1+dfrac11+x_1=dfrac32$ the positive number $x_1$ is the positive root of $$2X^2-X-1=0Rightarrowx_1=1,colorred text absurde$$
A fortiori for $x_1+dfrac11+x_1gtdfrac32$ which end the proof.

Can someone from this get the general answer for $ngt2$?