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?
Best Answer
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 $$1\geq 2\left(1-\sum_i{\tau_i\over m_i}\right)>2\cdot{1\over2}=1\ .$$