The question is
For any real numbers $4a<2b<c$, does there always exists a real number $t$ such that $\{ta\},\{tb\},\{tc\}$ all lie in $(1/3,2/3]$? Where $\{x\}=x-\lfloor x\rfloor$ is the decimal part of $x$.
I have the following tries on it
- WLOG assume $a=1$, one can draw the region $\Omega$ of $(b,c)$ that guarantee the existence of such $t$. For any cubic, the religion corresponding is a hexagon. By drawing picture, one can see that if $a\in \Omega$ then $a/4\in \Omega$ (since $4\equiv 1\bmod 4$).
- I guess that such $t$ exists when and only when $abc=0$.
- I also guess that for any real numbers $a_1,\ldots,a_n$ there always exists a real number $t$ such that $\{ta_i\}\in (1/3,2/3]$ if and only if $a_1\ldots a_n\neq 0$. If $x\in \Omega$ and $0<c<1$ implies $cx\in \Omega$, then it follows not difficultly by induction on $n$.
- But one have reason that it does not hold for $n$ large, since the cubic $\prod (n_i+1/3,n_i+2/3]$ has probability of $1/3^n$ in whole space, when $n$ large, it is harder to see it cover complete 'field of vision'.
- I think it can also be solved by some theory (e.g. Diophantine approximate) like Kronecker's theorem in high dimension.