Why are some other files written that the interval is $ \left(\frac{1}{3},\frac{2}{3}\right]$ not $ \left(\frac{1}{2},\frac{2}{3}\right]$? And what is the correct statement?

I checked the IMO Compendium, and it says that (1/3, 2/3] is the correct interval.

If $\lambda\in\left(\frac{3k+1}{3a}, \frac{3k+2}{3a}\right]$, then $\frac{1}{3}<\{\lambda a\}\leq \frac{2}{3}$. Define the intervals similarly for $b$ and $c$. Call the interval for $a$ - $L_k$, the interval for $b$ - $M_k$ and the interval for $c$ - $N_k$. Now the length of each $M_k$ is $\frac{1}{3b}$ which are spaced $\frac{2}{3b}$ apart from each other. The interval for each $N_k$ is $\frac{1}{3c}$ and they are spaced $\frac{2}{3c}<\frac{1}{3b}$ apart. Thus no $M_k$ can not contain an interval $N_k$ because the intervals not containing $N_k$ are all of length less than $\frac{1}{3b}$. Thus each interval $M_k$ must intersect some interval $N_k$. Now what remains to prove find is some interval $M_k$ that is contained entirely within some interval $L_k$. If that is true, then we want some $x$ and $y$ such that \[\frac{2x+1}{3a}\leq \frac{3y+1}{3b}<\frac{3y+2}{3b}\leq \frac{3x+2}{3a}.\] That is equivalent to finding some integer $x$ and $y$ such that \[\frac{3x+1}{3y+1}\leq \frac{a}{b}\leq \frac{3x+2}{3y+2}.\] Now if $\frac{a}{b}\leq \frac14$, then let $y=2^{e}$ where $e$ is the maximal $e$ such that $\frac{1}{3\cdot 2^e+1}\leq \frac{a}{b}$. The minimal $e$ is $0$ and thus the minimum $\frac{a}{b}$ is $\frac{1}{4}$ which explains the bound. Now then $\frac{1}{3\cdot 2^{e-1}+1}\geq \frac{a}{b}$ by definition so $\frac{2}{3\cdot 2^e+2}\geq \frac{a}{b}$. Therefore, if we let $x=0$ and $y={2^e}$, we have \[\frac{2x+1}{3y+1}=\frac{1}{3\cdot 2^e+1}\leq \frac{a}{b}\leq \frac{2}{3\cdot 2^e+2}=\frac{3x+2}{3y+2}.\] Now if $\frac{a}{b}\geq \frac{1}{4}$. Note that $\frac{a}{b}<\frac12$ because $b>2a$. Now the sequence $\frac{3\cdot2^e-2}{3\cdot2^{e+1}-2}$ approaches $\frac12$ as $e$ approaches infinity. Now choose the maximal $e$ such that $\frac{3\cdot2^e-2}{3\cdot2^{e+1}-2}\leq \frac{a}{b}$. Then $\frac{3\cdot 2^{e+1}-2}{3\cdot 2^{e+2}-2}=\frac{3\cdot 2^e-1}{3\cdot 2^{e+1}-1}\geq \frac{a}{b}$. Thus let $x=2^{e}-1$ and $y=2^{e+1}-1$ and thus that satisfies \[\frac{3x+1}{3y+1}\leq \frac{a}{b}\leq \frac{3x+2}{3y+2}.\] Therefore it is always possible to find some integer $x,y$ such that $M_k$ is contained entirely within some $L_k$, and the $M_k$ will always intersect some $N_k$ which proves that there exists some real number $\lambda$ that is in all three intervals so the fractional parts of $\lambda a, \lambda b,$ and $\lambda c$ are all in the interval $\left(\frac13, \frac23\right]$. $\blacksquare$

JSGandora wrote: That is equivalent to finding some integer $x$ and $y$ such that \[\frac{3x+1}{3y+1}\leq \frac{a}{b}\leq \frac{3x+2}{3y+2}.\] Alternatively, note that this is equivalent to $\frac{1}{3} \le \frac{ya - xb}{b-a} \le \frac{2}{3}$. If $a=gu$, $b=gv$ with $(u,v)=1$, we then just need to find $k\ge1$ with $\frac{1}{3} \le \frac{kg}{b-a} = \frac{k}{v-u} \le \frac{2}{3}$ (since $g = na-mb$ for some positive integers $n,m$). But $b>2a\implies v>2u$, so $v-u \ge u+1 \ge 2$, and we can simply take $k = \lfloor{(v-u)/2}\rfloor$ to finish (the "hardest" case is when $v-u=3$). I'm not sure if we can use rational approximation when $a/b$ is irrational (the problem still holds in this case by the previous solution), but at least the two-dimensional version of Kronecker's theorem doesn't seem to work directly.

Consider a "good $x$ interval" a interval of continuous $\lambda$s that work for $x$. It measures $1/3x$, and by inequality, any good-b interval intersects some good-c interval, so is enough to prove some good-b interval is completely inside a good-a interval. Is trivial to see that the $(\lambda_1+1)/2$'th good-b interval and the $(\lambda_2+1)/2$'th good-a interval satisfy this iff $ \frac{ | \lambda_1b - \lambda_2a | }{2} \le \frac{b-a}{3}$. By bezout lemma, is easy to find odd $\lambda$s that satisfy this (one needs to divide in cases $a|b$ and $a$ doesnt divide $b$), so we finish.

Solved with Smileyklaws Consider all intervals of the form $\left(\frac{3k+1}{3b}, \frac{3k+2}{3b}\right]$ for $k \ge 0$. Each of these intervals has length $\frac{1}{3b}$. Now consider all intervals of the form $\left(\frac{3k+1}{3c}, \frac{3k+2}{3c}\right]$ for $k \ge 0$. Each of these intervals has length $\frac{1}{3c}$, and the space between two consecutive intervals is $\frac{2}{3c}$. But $c > 2b \implies \frac{2}{3c} < \frac{1}{3b}$, so it follows that each $c$-interval overlaps with a $b$-interval. We wish to show that there exists some $\lambda$ in an $a$-interval, a $b$-interval, and a $c$-interval. Now we claim that there exists a $c$-interval that is contained within an $a$-interval. Note that each $c$-interval has length $\frac{1}{3c}$, while the space between two consecutive intervals is $\frac{2}{3c}$. Then $$c>4a \implies \frac{4}{3c}<\frac{1}{3a} \implies \frac{1}{3c}+\frac{2}{3c}+\frac{1}{3c} < \frac{1}{3a}$$so with some effort it follows that there exists a $c$-interval contained within an $a$-interval. Since each $c$-interval overlaps with a $b$-interval, by our work above, there exists an $a$-interval, a $b$-interval, and a $c$-interval that overlap. Consider some real $d$ in the union. Then for some positive integer $m$, $$\frac{3m+1}{3a} < \lambda \le \frac{3m+2}{3a} \implies m+\frac{1}{3} < \lambda a \le m+\frac{2}{3} \implies \frac{1}{3} < \{\lambda a\} \le \frac{2}{3}$$Similarly, $\frac{1}{3} < \{\lambda b\} \le \frac{2}{3}$, $\frac{1}{3} < \{\lambda c\} \le \frac{2}{3}$. This is what we needed, so we are done. $\blacksquare$

JSGandora wrote: If $\lambda\in\left(\frac{3k+1}{3a}, \frac{3k+2}{3a}\right]$, then $\frac{1}{3}<\{\lambda a\}\leq \frac{2}{3}$. Define the intervals similarly for $b$ and $c$. Call the interval for $a$ - $L_k$, the interval for $b$ - $M_k$ and the interval for $c$ - $N_k$. Now the length of each $M_k$ is $\frac{1}{3b}$ which are spaced $\frac{2}{3b}$ apart from each other. The interval for each $N_k$ is $\frac{1}{3c}$ and they are spaced $\frac{2}{3c}<\frac{1}{3b}$ apart. Thus no $M_k$ can not contain an interval $N_k$ because the intervals not containing $N_k$ are all of length less than $\frac{1}{3b}$. Thus each interval $M_k$ must intersect some interval $N_k$. Now what remains to prove find is some interval $M_k$ that is contained entirely within some interval $L_k$. If that is true, then we want some $x$ and $y$ such that \[\frac{2x+1}{3a}\leq \frac{3y+1}{3b}<\frac{3y+2}{3b}\leq \frac{3x+2}{3a}.\]That is equivalent to finding some integer $x$ and $y$ such that \[\frac{3x+1}{3y+1}\leq \frac{a}{b}\leq \frac{3x+2}{3y+2}.\]Now if $\frac{a}{b}\leq \frac14$, then let $y=2^{e}$ where $e$ is the maximal $e$ such that $\frac{1}{3\cdot 2^e+1}\leq \frac{a}{b}$. The minimal $e$ is $0$ and thus the minimum $\frac{a}{b}$ is $\frac{1}{4}$ which explains the bound. Now then $\frac{1}{3\cdot 2^{e-1}+1}\geq \frac{a}{b}$ by definition so $\frac{2}{3\cdot 2^e+2}\geq \frac{a}{b}$. Therefore, if we let $x=0$ and $y={2^e}$, we have \[\frac{2x+1}{3y+1}=\frac{1}{3\cdot 2^e+1}\leq \frac{a}{b}\leq \frac{2}{3\cdot 2^e+2}=\frac{3x+2}{3y+2}.\]Now if $\frac{a}{b}\geq \frac{1}{4}$. Note that $\frac{a}{b}<\frac12$ because $b>2a$. Now the sequence $\frac{3\cdot2^e-2}{3\cdot2^{e+1}-2}$ approaches $\frac12$ as $e$ approaches infinity. Now choose the maximal $e$ such that $\frac{3\cdot2^e-2}{3\cdot2^{e+1}-2}\leq \frac{a}{b}$. Then $\frac{3\cdot 2^{e+1}-2}{3\cdot 2^{e+2}-2}=\frac{3\cdot 2^e-1}{3\cdot 2^{e+1}-1}\geq \frac{a}{b}$. Thus let $x=2^{e}-1$ and $y=2^{e+1}-1$ and thus that satisfies \[\frac{3x+1}{3y+1}\leq \frac{a}{b}\leq \frac{3x+2}{3y+2}.\]Therefore it is always possible to find some integer $x,y$ such that $M_k$ is contained entirely within some $L_k$, and the $M_k$ will always intersect some $N_k$ which proves that there exists some real number $\lambda$ that is in all three intervals so the fractional parts of $\lambda a, \lambda b,$ and $\lambda c$ are all in the interval $\left(\frac13, \frac23\right]$. $\blacksquare$ I think you meant $3x+1$ in LHS of the inequality instead of $2x+1$.

Note that $\{\lambda x\}\in \left(\tfrac {1}{3}, \tfrac {2}{3} \right]$ is equivalent to $\lambda \in \left(\tfrac{3k+1}{3x},\tfrac{3k+2}{3x}\right]$ for some integer $k$. Call these the $x$-intervals, for $x=a,b,c$. Each $x$-interval has a starting point of $\tfrac{3k+1}{3x}$, a length of $\tfrac{1}{3x}$ and are $\tfrac{2}{3x}$ apart. Since $\tfrac{2}{3c}<\tfrac{1}{3b}$, each $b$-interval intersects some $c$-interval at least somewhere. $~$ It remains to show that some $b$-interval is contained in another $a$-interval. To do this, some $b$-interval's starting point must be in \[\left(\frac{k}{a}+\frac{1}{3a},\frac{k}{a}+\frac{2}{3a}-\frac{1}{3b}\right] = \left(\frac{3bk+b}{3ab},\frac{3bk+2b-a}{3ab}\right]\]Meanwhile, the starting points of the $b$-intervals are \[\frac{1}{3b}+\frac{l}{b}=\frac{3la+a}{3ab}\]Clearly, if $b>4a$ then $2b-a-b>3a$ so there must exist some $l$ that makes $3la+a$ fall into the interval $(3bk+b,3bk+2b-a]$. If $b=4a$ then $7a$ and $2b-a$ coincide. $~$ Now, we claim that no matter what $r=\tfrac{b}{a}$ is, we will be able to find a value $k$ and $l$. We have already done $k\ge 4$. Consider $l=2k+1$. We have $3la+a=(6k+4)a$. As long as \[b(3k+1) < a(6k+4) \le b(3k+2)-a \implies 2+\frac{1}{3k+2}\le r < 2+\frac{2}{3k+1}\]Thus, it is true for $r\in[2+\tfrac1{3k+2},2+\tfrac2{3k+1})$. Note that the lower bound can get arbitrarily close to $2$. It remains to show that the upper bound is at least equal to the lower bound of the previous, that for all $k\ge 1$, \[\frac2{3k+1}\ge \frac1{3k-1}\]Which is true. Note that when $k=0$ the upper bound is $4$. Therefore, if $r\in (2,4)$ we can find a value of $k,l$ that fit. We are done.

onk b-interval intersects c-interval prove b-interval is contained by a-interval now do spam stuff (oops i actually goofed and got to @above for $k=0$ and didn't generalize L me) wait just do what gandora did trust okay yeah cool my brain just farted but this works. hopefully i don't forget this but the moral of the story is to just do it smh my head

Define $R_a$ to be the union of all intervals $\left( \frac{3m+1}{3a}, \frac{3m+2}{3a} \right]$ where $m$ is an integer. Define $R_b, R_c$ similarly. Clearly $\{a\} \in \left(\frac {1}{3}, \frac {2}{3} \right]$ if and only if $\lambda \in R_a,$ and the same holds for $b, c.$ Therefore it suffices to show that $R_a \cap R_b \cap R_c$ is nonempty. We first show that every interval in $R_b$ intersects some $R_c.$ Note that each interval in $R_c$ has a length of $\frac{1}{3c}$ and are spaced $\frac{2}{3c}$ apart. The same applies to the intervals in $R_b.$ However, note that $\frac{2}{3c} < \frac{1}{3b},$ therefore for each interval in $R_b$ eventually one interval in $R_c$ will intersect it as they "shift" across the number line. To finish, it suffices to show that there is some interval in $R_b$ that lies completely within one in $R_a.$ To do this, we have to show that there exists integers $x, y$ such that $$\frac{3x+1}{3a} \leq \frac{3y+1}{3b} \text{ and } \frac{3x+2}{3a} \geq \frac{3y+2}{3b}.$$This is equivalent to $$\frac{3x+1}{3y+1} \leq \frac{a}{b} \leq \frac{3x+2}{3y+2}.$$Then, we can take cases based on $a, b$'s residues mod $3$ manually and finish. QED

Let $A_i = \left(\frac{i+1/3}{a}, \frac{i+2/3}{a}\right]$, and define $B_i$ and $C_i$ similarly (where $i$ is a nonnegative integer). First we claim that $B_j\subseteq A_i$ for some $i, j$. If $\frac{b}{a}> 4$, then take some integer $j$ from $\left[\frac{b/a-1}{3},\frac{2(b/a-1)}{3}\right]$ (which exists, given that the length of the interval is greater than $1$), and let $i=0$. Then $$\frac{i+1/3}{a}\le \frac{j+1/3}{b}<\frac{j+2/3}{b}\le \frac{i+2/3}{a},$$hence $B_j\subseteq A_i$. Otherwise, $2<\frac{b}{a}\le 4$. Let $I_n=\left[\frac{6n+5}{3n+2},\frac{6n+4}{3n+1}\right]$ for nonnegative integers $n$. Notice that $I_n$ and $I_{n+1}$ intersect, and $\frac{6n+5}{3n+1}$ tends to $2$ for large $n$. This implies that $I_0\cup I_1\cup I_2\cup\dots=\left(2,4\right]$, so $\frac{b}{a}\in I_n$ for some $n$. This implies $$\frac{n+1/3}{a}\le \frac{(2n+1)+1/3}{b}<\frac{(2n+1)+2/3}{b}\le \frac{n+2/3}{a},$$so taking $i=n$ and $j=2n+1$ suffices for $B_j\subseteq A_i$. Next, we claim that $B_j$ intersects some $C_k$. Assume otherwise. As $C_0=\left(\frac{1}{3c}, \frac{2}{3c}\right]$, $B_j=\left(\frac{j+1/3}{b},\frac{j+2/3}{b}\right]$ and $\frac{j+1/3}{b}>\frac{1}{3c}$, each element of $C_0$ is strictly less than each element of $B_j$. Now let $m$ be the largest nonnegative integer for which each element of $C_m$ is strictly less than each element of $B_j$. Then each element of $C_{m+1}$ is strictly greater than each element of $B_j$. In light of these facts, for arbitrary $\delta > 0$ we have $$\frac{m+2/3}{c}<\frac{j+1/3}{b}+\delta\quad\text{and}\quad \frac{j+2/3}{b}<\frac{(m+1)+1/3}{c}+\delta,$$so subtracting these inequalities yields $\frac{1}{3b}<\frac{2}{3c}+2\delta$. Taking $\delta=\frac{1}{2}\left(\frac{1}{3b}-\frac{2}{3c}\right)>0$, we arrive at a contradiction. Finally, there is some $\lambda\in B_j, C_k$. By construction, $\lambda\in A_i$ as well. Now $\lambda a\in \left(i+\frac{1}{3},i+\frac{2}{3}\right]$ and similarly for $b$ and $c$, as desired.

i died bounding only to find the same solution as #4