Let $n \geq 2$ be an integer. Positive reals $x_1, x_2, \cdots, x_n$ satisfy $x_1x_2 \cdots x_n = 1$. Show: $$\{x_1\} + \{x_2\} + \cdots + \{x_n\} < \frac{2n-1}{2}$$Where $\{x\}$ denotes the fractional part of $x$.
Problem
Source: CWMI 2018 Q2
Tags: algebra, inequalities, fractional part, n-variable inequality
TuZo
15.08.2018 09:10
I think that this is easy, because the induction step is: $\frac{2n-1}{2}+\left\{ {{x}_{n+1}} \right\}<\frac{2n+1}{2}\Leftrightarrow \left\{ {{x}_{n+1}} \right\}<1\text{ it is true}$
taiat
15.08.2018 11:52
TuZo wrote: I think that this is easy, because the induction step is: $\frac{2n-1}{2}+\left\{ {{x}_{n+1}} \right\}<\frac{2n+1}{2}\Leftrightarrow \left\{ {{x}_{n+1}} \right\}<1\text{ it is true}$ it's wrong to use mathematical induction by this way，you ignore that x1*x2*...xn=1.
TuZo
15.08.2018 11:58
Yes indeed, the corrected induction step is: $\frac{2n-1}{2}+\left\{ {{x}_{n}} \right\}+\left\{ {{x}_{n+1}} \right\}-\left\{ {{x}_{n}}{{x}_{n+1}} \right\}\le \frac{2n+1}{2}$, but this is evident true.
taiat
15.08.2018 12:07
TuZo wrote: Yes indeed, the corrected induction step is: $\frac{2n-1}{2}+\left\{ {{x}_{n}} \right\}+\left\{ {{x}_{n+1}} \right\}-\left\{ {{x}_{n}}{{x}_{n+1}} \right\}\le \frac{2n+1}{2}$, but this is evident true. The induction works only if x1*x2*...xn=1，but x1*x2*...*xn+1=1 ，it shows that xn+1=1.
TuZo
15.08.2018 12:34
No, no work fine for all situation, here is the prove: We suppose that for all $x_1, x_2, \cdots, x_n$ satisfy $x_1x_2 \cdots x_n = 1$ result $\{x_1\} + \{x_2\} + \cdots + \{x_n\} < \frac{2n-1}{2}$. Now we consider the ${{x}_{1}},{{x}_{2}},...,{{x}_{n}},{{x}_{n+1}}\text{ with }{{x}_{1}}{{x}_{2}}...{{x}_{n}}{{x}_{n+1}}=1\Leftrightarrow {{x}_{1}}{{x}_{2}}...{{x}_{n-1}}\left( {{x}_{n}}{{x}_{n+1}} \right)=1$, and here we have $n$ numbers with the product $1$, so it is true that $\{{{x}_{1}}\}+\{{{x}_{2}}\}+\cdots +\{{{x}_{n-1}}\}+\left\{ {{x}_{n}}{{x}_{n+1}} \right\}<\frac{2n-1}{2}\Leftrightarrow \{{{x}_{1}}\}+\{{{x}_{2}}\}+\cdots +\{{{x}_{n}}\}+\left\{ {{x}_{n+1}} \right\}\le \frac{2n-1}{2}+\{{{x}_{n}}\}+\left\{ {{x}_{n+1}} \right\}-\left\{ {{x}_{n}}{{x}_{n+1}} \right\}$, so on the basis of the induction it is enought to prove that $\frac{2n-1}{2}+\left\{ {{x}_{n+1}} \right\}<\frac{2n+1}{2}\Leftrightarrow \left\{ {{x}_{n+1}} \right\}<1\text{ it is true}$, what I did in my post #3!
primes020
15.08.2018 16:50
TuZo wrote: Yes indeed, the corrected induction step is: $\frac{2n-1}{2}+\left\{ {{x}_{n}} \right\}+\left\{ {{x}_{n+1}} \right\}-\left\{ {{x}_{n}}{{x}_{n+1}} \right\}\le \frac{2n+1}{2}$, but this is evident true. $\frac{2n-1}{2}+\left\{ {{x}_{n}} \right\}+\left\{ {{x}_{n+1}} \right\}-\left\{ {{x}_{n}}{{x}_{n+1}} \right\}\le \frac{2n+1}{2}$ isn't always correct take ${x}_{n}=0.9 , {x}_{n+1}=1.9$
primes020
15.08.2018 17:29
I'm using phone so I'll only write a sketch. First assume the inequality isn't right and WOLG $x_1 \leq x_2 \dots \leq x_n$ and $x_{k}$ is the largest number less than 1 we have $\sum_{i=1}^{k} x_i \geq \frac{2k-1}{2}$ and $x_{k+i} \geq 1.5$ use the inequality $a_1 a_2...a_n +n-1> a_1+a_2...+a_n(0<a_i<1)$ $x_1 x_2 ... x_k > \frac{1}{2}$ because $x_1... x_n =1$, $x_{k+1}...x_n <2$ but $x_{k+i} \geq 1.5$ so only $x_n>1$. we only need to prove $x_1 x_2 \dots x_{n-1} + n-1+x_n-[x_n]<\frac{2n-1}{2}$ this is the "$n=2$" case
TuZo
15.08.2018 18:05
primes020 wrote: TuZo wrote: Yes indeed, the corrected induction step is: $\frac{2n-1}{2}+\left\{ {{x}_{n}} \right\}+\left\{ {{x}_{n+1}} \right\}-\left\{ {{x}_{n}}{{x}_{n+1}} \right\}\le \frac{2n+1}{2}$, but this is evident true. $\frac{2n-1}{2}+\left\{ {{x}_{n}} \right\}+\left\{ {{x}_{n+1}} \right\}-\left\{ {{x}_{n}}{{x}_{n+1}} \right\}\le \frac{2n+1}{2}$ isn't always correct take ${x}_{n}=0.9 , {x}_{n+1}=1.9$ Yes, you are right, so the induction it do not work!
taiat
16.08.2018 13:33
TuZo wrote: Yes indeed, the corrected induction step is: $\frac{2n-1}{2}+\left\{ {{x}_{n}} \right\}+\left\{ {{x}_{n+1}} \right\}-\left\{ {{x}_{n}}{{x}_{n+1}} \right\}\le \frac{2n+1}{2}$, but this is evident true. in fact we don't need to control LHS very well，by Bernoulli inequality we have x1+x2+...+xk≤k-1+x1*x2..*xk（we amuse x1，x2...xk≤1），it's enough to prove the inequality.
RagvaloD
16.08.2018 18:48
Solution follows from induction and next lemma: If $a <1,ab<1$ then $\{a\}+\{b\}-\{ab\}<1$ Proof: If $a < 1,b<1$ then $\{a\}+\{b\}-\{ab\}-1=a+b-ab-1=-(1-a)(1-b)<0$ If $a<1,b>1,ab<1$ then $\{a\}+\{b\}-\{ab\}-1=a+b-ab-1-[b]=-[b]-(a-1)(b-1)<-(b-1)-(a-1)(b-1)=-a(b-1)<0$
WAit_Mng
09.07.2020 17:37
Could someone check the following please?
We induct on $n$.
When $n=2$, it's trivial.
Assume the inequalitty holds for $n$, we prove it true for $n+1$ $x_i$'s with $x_1x_2\ldots x_{n+1}=1$
If there exists $x_n<1$ and $x_{n+1}<1$
Note that $\{{{x}_{1}}\}+\{{{x}_{2}}\}+\cdots +\{{{x}_{n-1}}\}+\left\{ {{x}_{n}}{{x}_{n+1}} \right\}<\frac{2n-1}{2}$.
So it suffices to show $x_n + x_{n+1}-x_nx_{n+1} <1$, which is obvious.
Now if there is only one $x_i$ such that $x_i<1$. We show that there must exist $\{x_k\}<\frac{1}{2}$, from which our original inequality holds.
Assume to the contrary that all terms are greater than $1/2$, then since there is only one $x_i<1$, $x_1x_2\ldots x_{n+1}\geq \frac{1}{2} (\frac{3}{2})^{n-1}>1$, contradiction. So there exists $\{x_k\}<\frac{1}{2}$.
Thus, the inequality holds true for $n+1$.
SMSGodslayer
20.09.2021 15:23
WAit_Mng wrote: Could someone check the following please?
We induct on $n$.
When $n=2$, it's trivial.
Assume the inequalitty holds for $n$, we prove it true for $n+1$ $x_i$'s with $x_1x_2\ldots x_{n+1}=1$
If there exists $x_n<1$ and $x_{n+1}<1$
Note that $\{{{x}_{1}}\}+\{{{x}_{2}}\}+\cdots +\{{{x}_{n-1}}\}+\left\{ {{x}_{n}}{{x}_{n+1}} \right\}<\frac{2n-1}{2}$.
So it suffices to show $x_n + x_{n+1}-x_nx_{n+1} <1$, which is obvious.
Now if there is only one $x_i$ such that $x_i<1$. We show that there must exist $\{x_k\}<\frac{1}{2}$, from which our original inequality holds.
Assume to the contrary that all terms are greater than $1/2$, then since there is only one $x_i<1$, $x_1x_2\ldots x_{n+1}\geq \frac{1}{2} (\frac{3}{2})^{n-1}>1$, contradiction. So there exists $\{x_k\}<\frac{1}{2}$.
Thus, the inequality holds true for $n+1$.
$x_1x_2\ldots x_{n+1}\geq \frac{1}{2} (\frac{3}{2})^{n-1}>1$ should be $x_1x_2\ldots x_{n+1}\geq \frac{1}{2} (\frac{3}{2})^{n}>1$ for $n \geq {2}$