Assume otherwise. Then, if $A$ consists of $0 \leq x_1<x_2<\ldots<x_{4n+2} \leq 5^n$, then $3x_{4n+1} \geq x_{4n+2}+2x_{4n}$. Let $y_i=x_{4n+2}-x_i>0$ for all $i$, and so the above relation rewrites as $y_{i+1} \leq \frac{2}{3}y_i$. Hence, we easily obtain that $y_{4n+1} \leq (\frac{2}{3})^{4n} y_1$. However, $1 \leq y_{4n+1} \leq (\frac{2}{3})^{4n}(x_{4n+2}-x_1) \leq (\frac{16}{81})^n(5^n-0)=(\frac{80}{81})^n<1,$ which gives the desired contradiction.

Suppose the elements of $A$ are $a_1 < a_2 < \cdots < a_{4n+2}$. Pick $c = a_{4n+2}$ (with the idea in mind that we do not win more if we do not pick the largest element of $A$) and aim for $a$ and $b$ to be consecutive in the above sequence (in any other case the difference $3b-2a$ shall be larger). If we prove that $c + 2a_i > 3a_{i+1}$ for at least one $i=1,2,\ldots,4n$, the problem shall be solved. Suppose the contrary, i.e. $a_{i+1} \geq \frac{c+2a_i}{3}$ for all $i=1,2,\ldots,4n$. Setting $a_i = b_i + c$ yields $b_{i+1} \geq \frac{2b_i}{3}$ and so $b_{4n+1} \geq \left(\frac{2}{3}\right)^{4n}b_1$, whence $\left(\frac{2}{3}\right)^{4n}(c-a_1) \geq c - a_{4n+1}$. In the latter the right-hand side is at least $1$, while the left-hand one is less than $\left(\frac{2}{3}\right)^{4n} \cdot 5^n = \left(\frac{80}{81}\right)^{n} < 1$, contradiction.

ISL 2021 A1. Let $n\in\mathbb Z$ and $A\subseteq\{0, 1, \dots, 5^n\}$ with $|A|=4n+2$. Prove that there exist $a, b, c\in A$ such that $a<b<c$ and $c+2a>3b$. Solution. We choose elements of $A$ one by one from $\{0, 1, \dots, 5^n\}$ such that there are no $a, b, c\in A$ and $a<b<c$ such that $c+2a>3b$. Consider the case when $0$ and $5^n$ are chosen at the moment. Take $a=0$, $c=5^n$, then all the numbers less than $\left[\frac{5^n}3\right]+1$ cannot be chosen, so the smallest number can be chosen is greater than $\frac13\cdot 5^n$. Take $a$ be that number, which is greater than $\frac13\cdot 5^n$ and $c=5^n$, then all the numbers less than $\left[\frac{5^n+\frac23\cdot 5^n}3\right]+1$ cannot be chosen, so the smallest number can be chosen this time is greater than $\frac59\cdot 5^n$. The idea is to repeat this for $4n$ times. Define a sequence by $a_1=\frac13$ and $a_{n+1}=\frac{2a_n+1}3$ for all $n\in\mathbb Z_+$. It follows that the smallest number can be chosen in the $k^{\text{th}}$ is greater than $a_k\cdot 5^n$. Claim 1. We have $a_{4n}>\frac{5^n-1}{5^n}$. Proof. Induction. For the base case: $a_1=\frac13$, $a_2=\frac59$, $a_3=\frac{19}{27}$, $a_4=\frac{65}{81}>\frac45$. Assume that $a_{4k}>\frac{5^k-1}{5^k}$. Then \begin{align*}a_{4(k+1)}&=\frac{2\left(\frac{2\left(\frac{2\left(\frac{2a_{4k}+1}3\right)+1}3\right)+1}3\right)+1}3\\&=\frac{16a_{4k}+65}{81}\\&>\frac{16\cdot\frac{5^k-1}{5^k}+65}{81}\\&=\frac{81\cdot 5^k-16}{81\cdot 5^k}\\&>\frac{81\cdot 5^k-16-5^k}{81\cdot 5^k-5^k}\\&=\frac{5^{k+1}-1}{5^{k+1}}\end{align*}does the inductive step. It follows from Claim 1 that the smallest number can be chosen in the last time, i.e. the $4n^{\text{th}}$ time, is greater than $\frac{5^n-1}{5^n}\cdot 5^n=5^n-1$, but the only such number is available is $5^n$, which is chosen in the beginning. This is a contradiction. If the numbers $0$ or $5^n$ is not chosen, then repeat the argument by taking $a$ to be the minimal element of $A$ and $c$ the maximal element of $A$. The contradiction above will still be induced as the range of 'available numbers' is even smaller than between $1$ and $5^n-1$ inclusive in above. Therefore, there is no way we can choose numbers such that there are no $a, b, c\in A$ and $a<b<c$ such that $c+2a>3b$, the result follows.

Let $A=\{a_1,a_2,\ldots,a_{4n+2}\}$ with $a_1<a_2<\ldots<a_{4n+2}$. If $a$, $b$, and $c$ do not exist, we must have that $a_{4n+2}+2a_i\leq3a_{i+1}$ for all $1\leq i\leq4n+1$. We claim that $a_i\geq\left(1-\left(\frac23\right)^{i-1}\right)a_{4n+2}$ by induction on $i$. Base Case: $i=1$ Since $a_1\geq0$, we have $a_1\geq\left(1-\left(\frac23\right)^{1-1}\right)a_{4n+2}=0$. Inductive Step: We have \begin{align*} a_{i+1}\geq\frac13a_{4n+2}+\frac23a_i\\ &\geq\frac13a_{4n+2}+\frac23a_{4n+2}-\left(\frac23\right)^ia_{4n+2}\\ &=\left(1-\left(\frac23\right)^i\right)a_{4n+2}. \end{align*}Therefore, $a_{4n+1}\geq\left(1-\left(\frac23\right)^{4n}\right)a_{4n+2}>\left(1-\left(\frac15\right)^n\right)a_{4n+2}$ since $\left(\frac23\right)^4=\frac{16}{81}<\frac15$. Since $a_{4n+2}\leq5^n$, we have $a_{4n+1}>a_{4n+2}-\frac{a_{4n+2}}{5^n}\geq a_{4n+2}-1$, which is impossible since $a_{4n+1}\leq a_{4n+2}-1$. Therefore, this is a contradiction, so there must exist $a$, $b$, $c\in A$ such that $a<b<c$ and $c+2a>3b$.

Suppose for the sake of contradiction that $c+2a\le 3b$ for all $a,b,c\in A$ with $a<b<c.$ $\textbf{Claim: }$ If the elements of $A$ are $s_1<s_2<\dots<s_{4n+2},$ then $s_i\ge s_{4n+2}(1-(\tfrac{2}{3})^{i-1})$ for all $i.$ $\emph{Proof: }$ Induct on $i;$ the base case is true because $s_1\ge 0.$ For the inductive step, consider $s_k.$ Take $(a,b,c)=(s_{k-1},s_k,s_{4n+2});$ by our assumption, we have \[s_{4n+2}+2s_{k-1}\ge 3s_k.\]Therefore, \begin{align*} s_k &\ge\frac{s_{4n+2}+2s_{k-1}}{3}\\ &\ge\frac{s_{4n+2}+2(s_{4n+2}(1-(\tfrac{2}{3})^{k-2}))}{3}\\ &=\frac{s_{4n+2}(3-2(\tfrac{2}{3})^{k-2})}{3}\\ &=s_{4n+2}(1-(\tfrac{2}{3})^{k-1}), \end{align*}as desired. $\blacksquare$ In particular, we have \begin{align*} s_{4n+1} &\ge s_{4n+2}(1-(\tfrac{2}{3})^{4n})\\ &=s_{4n+2}-(\tfrac{16}{81})^{n}s_{4n+2}\\ &>s_{4n+2}-(\tfrac{1}{5})^{n}5^n\\ &=s_{4n+2}-1. \end{align*}This is impossible, as $s_{4n+1}$ is an integer less than $s_{4n+2}.$

The idea is to plug in triples of the form $(a_i, a_{i+1}, a_{4n+2})$ (of course, let $a_1<...<a_{4n+2}$). Supposing otherwise for contradiction, we get $3a_{i+1} \geq 2a_i+a_{4n+2}$, so if $d_i=a_i-a_{4n+2}$, then $3d_{i+1} \geq 2d_i \iff \frac {d_{i+1}} {d_i} \geq \frac {2}{3}$, so $d_{4n+1} \geq (\frac{2}{3})^{4n}d_1$ by multiplying them. Then $(5.(\frac{2}{3})^4)^n>(\frac{2}{3})^{4n}(a_{4n+2}-a_1) \geq a_{4n+2}- a_{4n+1}\geq 1$, contradiction.

Order the elements $x_1 < x_2 < x_3 < \cdots < x_{4n+1} < x_{4n+2}$. We claim that for some $1 \le i \le 4n$, $(a_i,a_{i+1},a_{4n+2})$ satisfies the condition. Assume not. Let $d_i=x_{4n+2}-x_i$. The condition becomes $d_{i+1} \le \frac{2}{3} d_i$. This means that $d_{4n+1} \le \left( \frac{2}{3} \right) ^{4n} d_1$. We have that $d_{4n+1} \ge 1$ and $d_1 \le 5^n$. We get that $$1 \le d_{4n+1} \le \left( \frac{2}{3} \right) ^{4n} d_1 \le \left( \frac{2}{3} \right) ^{4n} \cdot 5^n = \left( \frac{80}{81} \right) ^n,$$which is an obvious contradiction so we are done.

Let the elements in $A$ be $a_1 < a_2 < \ldots < a_{4n+2}$. Suppose there exists a tuple $(i,j,k)$ of indices $i < j < k$ such that $a_{k} + 2a_{i} > 3a_j$. Then note that $a_{4n+2} + 2a_{j-1} \ge a_k + 2a_i > 3a_j$, so the tuple $(j-1,j,4n+2)$ also works. Therefore, it is equivalent to show that for some $1 \le j \le 4n-1$, we have $x_{4n+2} + 2x_{j-1} > 3x_j$. Assume that this is not the case. Then for all $1 \le j \le 4n-1$, we have \[2(x_{j-1} - x_{4n+2}) \le 3(x_{j} - x_{4n+2}) \implies x_{4n+2}-x_j \le \frac{2}{3} (x_{4n+2}-x_{j-1}) \]In particular, this means \[x_{4n+2} - x_{4n+1} \le \left ( \frac{2}{3} \right )^{4n} (x_{4n+2} - x_1) \le \left ( \frac{2}{3} \right )^{4n} (5^n - 0) = \left ( \frac{80}{81} \right )^n < 1,\]a contradiction.

A bit hastily written. Let the elements of $A$ be $a_1<a_2<\dots<a_{4n+2}$. Note that adding or subtracting any constant $C$ from each element in $A$ doesn't change whether it satisfies the condition in the problem; therefore, we may set $a_{4n+2}=5^n$. Note that the condition is equal to $c-b>2(b-a)$; it suffices to show that we can't construct a set such that $a_{4n+2}-a_{i+1}\le 2(a_{i+1}-a_i)$ for all $1\le i\le 4n$. Now let $a_{4n+2}-a_{4n+1}=d$. Then \[a_{4n+1}-a_{4n}\ge \frac{d}{2}\]\[a_{4n}-a_{4n-1}\ge \frac{3d}{4}\]\[\vdots\]\begin{align*} a_{i+1}-a_{i}&\ge \frac{a_{4n+2}-a_{i+1}}{2} \\ &=\frac{(a_{4n+2}-a_{4n+1})+(a_{4n+1}-a_{4n})+\dots+(a_{i+2}-a_{i+1})}{2} \\ &\ge \frac{d}{2}\cdot \left(\frac{3}{2}\right)^{4n-i}. \end{align*}Finally we obtain $5^n\ge a_{4n+2}-a_{1}\ge d\left(\frac{3}{2}\right)^{4n}\ge \left(\frac{81}{16}\right)^n>5^n$ which is a contradiction. We are done. $\blacksquare$

Nice, I like it! Let the numbers in $A$ be $a_1,a_2,\dots,a_{4n+2}$, where $a_i<a_j$ if $i<j.$ Assume FTSOC that $3b\ge 2a+c$ for all $a,b,c\in A.$ Thus, we get that $a_2\ge \frac{2a_1+a_{4n+2}}{3}\ge \frac{1}{3}a_{4n+2}.$ Call a positive integer $1\le k\le 4n+1$ $r$-good if we have $a_k\ge ra_{4n+2}.$ If $a_i$ is $r$-good then $a_{i+1}$ is $\frac{2r+1}{3}$-good. Note that $a_2$ is $1-\frac{2}{3}$-good. If $a_i$ is $1-\left(\frac{2}{3}\right)^{i-1}$-good then we see that $a_{i+1}$ is $1-\left(\frac{2}{3}\right)^i$-good. Thus, by induction, $a_{4n+1}$ is $1-\left(\frac{2}{3}\right)^{4n}$-good. Hence, we have $a_{4n+1}\ge a_{4n+2}-a_{4n+2}\cdot \left(\frac{16}{81}\right)^n.$ We claim that it is forced to have $a_{4n+1}=a_{4n+2}.$ To do this, we show that $a_{4n+2}\left(\frac{16}{81}\right)^n< 1.$ This is true, since $a_{4n+2}\left(\frac{16}{81}\right)^n< \frac{a_{4n+2}}{5^n}\le 1.$ Thus, we reach a contradiction, so we're done.

This problem was proposed by Burii and me

Let the subset contain elements $x_1<x_2<\dots<x_{4n+2}$ and let $c=x_{4n+2}.$ Suppose FTSOC that $x_{4n+2}+2a\le 3b$ for all $a<b,$ which implies $x_{4n+2}+2x_i\le 3x_{i+1}$ for all $1\le i\le 4n.$ We claim by induction that $x_j\ge x_{4n+2}(1-(2/3)^{j-1})$ for $j\ge 1.$ Indeed, $x_1\ge 0=x_{4n+2}(1-1)$ and $x_k\ge x_{4n+2}(1-(2/3)^{k-1})$ implies $$x_{k+1}\ge \frac{x_{4n+2}+2x_k}{3}\ge x_{4n+2}\frac{1+1-2(2/3)^j}{3}=x_{4n+2}(1-(2/3)^k),$$completing the induction. Notice $$x_{4n+1}\ge x_{4n+2}(1-(16/81)^n)>x_{4n+2}(1-(1/5)^n)=x_{4n+2}-x_{4n+2}(1/5)^n\ge x_{4n+2}-1$$since $1/5>16/81$ and $x_{4n+2}\le 5^n.$ Hence, $x_{4n+1}>x_{4n+2}-1$ so $x_{4n+1}\ge x_{4n+2},$ a contradiction. $\square$

Suppose that $c+2a \leq 3b$ for all $a < b < c$ in $A.$ Then $a \leq \frac{3b - c}{2}.$ Then $A = \{a_1,\ldots,a_{4n+2}\}$ where $a_1 < \cdots < a_{4n+2},$ we have for all $k \leq 4n$ and considering the triple $(a_k, a_{k+1}, a_{4n+2}),$ that $a_k \leq \frac{3 a_{k+1} - a_{4n+2}}{2}.$ Thus, let $x = a_{4n+2} \leq 5^n, y = a_{4n+1} \leq x-1.$ Then $a_{4n} \leq \frac{3y-x}{2},$ and $a_{4n-1} \leq \frac{3 \left( \frac{3y-x}{2} - x \right)}{2} = \frac{9y-5x}{4},$ and $a_{4n-2} \leq \frac{3 \left( \frac{9y-5x}{4} \right) - x}{2} = \frac{27y-19x}{8},$ etc. We can easily see that in general we have by induction $a_{4n+1-m} \leq \frac{3^m y - (3^m - 2^m) x}{2^m}$ by induction, as $\frac{ 3 \left( \frac{3^j y - (3^j - 2^j) x}{2^j} \right) - x }{2} = \frac{ 3^{j+1} y - 3(3^j - 2^j)x - 2^j x }{2^{j+1}} = \frac{3^{j+1} y - (3^{j+1} - 2^{j+1}) x}{2^{j+1}}$ Thus, we get $a_1 \leq \frac{3^{4n} y - (3^{4n} - 2^{4n}) x}{2^{4n}} \leq \frac{3^{4n} (x-1) - (3^{4n} - 2^{4n}) x}{2^{4n}} = \frac{-3^{4n} + 2^{4n} x}{2^{4n}} = x - \frac{3^{4n}}{2^{4n}} \leq 5^n - \frac{3^{4n}}{2^{4n}} < 0$ where the last inequality holds for all $n \geq 1$ as $5^n < \frac{81^n}{16^n}$ follows from $80^n < 81^n.$ This is a contradiction. $\blacksquare$

Fun problem

Assume there exist subset $A=\left \{ a_1,a_2,...,a_{4n+2} \right \}$ with $a_i\leq a_{i+1}$ which doesn't satisfy the condition. In particular, for every $1\leq k\leq 4n$ holds $$a_{4n+2}+2a_k\leq 3a_{k+1}\implies a_{4n+2}-a_k\geq \frac 32 (a_{4n+2}-a_{k+1})\geq (\frac 32)^2 (a_{4n+2}-a_{k+2})\geq ...\geq (\frac 32)^{4n+1-k}(a_{4n+2}-a_{4n+1}).$$For $k=1$ get the contradiction with $a_{4n+2}-a_1\geq (\frac{81}{16})^n (a_{4n+2}-a_{4n+1})>5^n$.

Posting for storage Let the elements of $A$ be $1 \le a_1 \le a_2 \le ... \le a_{4n+2} \le 5^n$. FTSOC, assume otherwise hence for any $a<b<c$ we have $$ c+2a \le 3b \Leftrightarrow \frac{ c+2a}{3} \le b$$Then this inequality is true in the "worst case" i.e $c= a_{4n+2}$ and $b,a$ are consecutive variables. Then we have $$ \frac{ a_{4n+2} +2a_i}{3} \le a_{i+1}$$Observe that we can reapply the same inequality on $2a_i$ or in other words, we will have a chain inequality and in result, we will have bounds between $a_1,a_{4n+1}$. Let $a_i+b_i= a_{4n+2}$. We have $$ \frac{2}{3} b_i \ge b_{i+1} \Rightarrow b_{4n+1} \le \left( \frac{2}{3} \right)^{4n} b_1$$But we have $1 \le b_{4n+1}$ and $b_1 \le 5^n$. Hence $$ 1 \le \left( \frac{2}{3} \right)^{4n}\cdot 5^n= \left( \frac{80}{81} \right)^n<1$$Contradiction, hence there exist such $a,b,c$ and we are done.

Solved for OTIS

Call a subset of $\{0, 1, \dots, 5^n\}$ a connected subset if, for all $a<b<c$, $c+2a\leq 3b$ . Consider the largest possible cardinality $z$ of a connected subset. We construct a set $B$ with $|B|=z$. We construct $B$ from its largest element to its smallest element. It is clearly optimal to add $5^n$ to $B$ (otherwise, simply increase all elements by an equal amount until the largest element is $5^n$, which still satisfies the desired property). At any given point $B'$, while constructing $B$, let the maximum element of $B'$ be $p$ (which is equal to $5^n$) and the minimum element of $B$ be $q$. Then for any integers $r,s\in B'$ with $r<s$, if we add $t<r$ to $B'$, we need $s+2t\leq 3r$, i.e., $t\leq\frac{3}{2}r-\frac{1}{2}s$. This value is minimized when $r=q$ and $s=p$. Hence, we can add any integer between $0$ and $\frac{3}{2}q-\frac{1}{2}p$, inclusive, and still have $c+2a\leq 3b$ for all $a<b<c$ in $B$. In fact, it is not hard to see that it is optimal to add the largest integer $x$ that is at most $\frac{3}{2}q-\frac{1}{2}p$ (if we do not do so, then every element added afterwards will be less than what was possible had we added $x$, resulting in a connected subset with cardinality at most what was possible if we had added $x$). Through a similar proof, it is also not hard to see that it is optimal to add $5^n-1$ into the set right after adding $5^n$. In each step, we update the minimum value of $B'$ but leave the maximum constant as $5^n$. Therefore, we have created a recursive upper bound for each element of the final result $B$: if the upper bound for the last element added was $a_i$, then the upper bound for the next element is $a_{i+1}=\frac{3}{2}a_i-\frac{5^n}{2}$. It is easily provable by induction that for the $i$th element of $B$, this upper bound is equal to $a_i=5^n-\left(\frac{3}{2}\right)^{i-2}$, as the first element of $B$ is $5^n$ and the second element is $5^n-1$. Therefore, as $|B|=z$, we must have $5^n-\left(\frac{3}{2}\right)^{z-2}\geq 0$. This can be rewritten as \[z\leq 2+n\log_{3/2}5.\]As $\left(\frac{3}{2}\right)^4=\frac{81}{16}>5$, we know that $\log_{3/2}5<4$, so $z<4n+2$. But this means that for any subset $A$ of $\{0, 1, \dots, 5^n\}$ with $4n+2$ elements, $A$ cannot be a connected subset, i.e., there must exist $a<b<c$ in $A$ with $c+2a>3b$, as desired.

Assume for the sake of contradiction that the condition is never true. Then, let the elements of $S$ be $a_1, a_2, \cdots, a_{4n+2}$ in increasing order. The condition then implies $$a_{4n+2}-a_2 \leq \frac 23(a_{4n+2}-a_1).$$Inductively, we have $$a_{4n+2} - a_{4n+1} \leq \left(\frac 23\right)^{4n}(a_{4n+2} - a_1) \leq \left(\frac 23\right)^{4n} \cdot 5^n.$$But as $5 \cdot \left(\frac 23\right)^4 < 1$, this implies $a_{4n+2} - a_{4n+1} < 1$, which is an obvious contradiction.

Let the elements be $x_1<x_2< \dots < x_{4n+2}$. Assume this doesn't hold. Then we can write \[ x_{4n+2} - x_i \leq \frac 23 (x_{4n+2} -x _{i-1}). \] It is easy to see that $x_{4n+2} - x_{4n+1} \leq \left( \frac{80}{81} \right)^n <1$, which is obviously false as the elements are integers.

Assume otherwise. Let $A=\{a_1<\dots<a_{4n+2}\}$; substituting $(a,b,c):=(a_k,a_{k+1},a_{4n+2})$ gives $2a_k+a_{4n+2}\le3a_{k+1}$ or $a_{4n+2}-a_{k+1}\le\tfrac{2}{3}(a_{4n+2}-a_k)$. It follows that \[a_{4n+2}-a_{4n+1}\le(\tfrac{2}{3})^{4n}(a_{4n+2}-a_1)\le(\tfrac{2}{3})^{4n}5^n=(\tfrac{80}{81})^n<1,\]contradiction. $\square$

Assume not. Let the elements be $a_1<a_2<\dots<a_{4n+2}$; we have that $a_{4n+2}-a_{k+1}\le\tfrac{2}{3}(a_{4n+2}-a_k)$, so simplifying this inequality-recurrence yields $$a_{4n+2}-a_{4n+1}\le\left(\frac 23\right)^{4n}(a_{4n+2}-a_1)<\left(\frac 23\right)^{4n}5^n=\left(\frac{80}{81}\right)^n<1,$$contradiction. $\blacksquare$ super easy problem

The condition $c+2a>3b$ with $a<b<c$ can be rewritten as $(c-b)>2(b-a)$ for where $c-b$ and $b-a$ are positive. FTSOC, assume that for some choice of $A$, we have $(c-b)\leq 2(b-a)$ for all $a,b,c\in A$. Then, out of all such $A$, there must exist some choice of $A$ with the smallest range. Let $5^n$ be the largest element of $A$, and we try to make each subsequent, smaller element of $A$ as large as possible. Our sequence becomes $5^n, 5^n-1, 5^n-2, 5^n-3, 5^n-5,\ldots$ where if some element is $5^n-k$, the following element is $5^n-\left(k+\left\lceil \frac{k}{2}\right\rceil\right)=5^n-\left\lceil\frac{3}{2}k\right\rceil$. That's because if we try to make $c-b$ as large as possible and $b-a$ small as possible, we must let $a$ and $b$ adjacent elements, and $c$ be the largest element, so we obtain $c-b=5^n-(5^n-k)=k\leq 2\cdot (b-a)$, meaning $\left\lceil\frac{k}{2}\right\rceil$ is the least possible difference between $b$ and $a$. We now know there exists such an $A$ if the $4n+2$th element is $5^n-k_{4n+2}$ and $k_{4n+2}<5^n$. However, we see that for the $i$th element, $5^n-k_i$, $k_i>\left(\frac{3}{2}\right)^{i-2}$, which can be easily shown by induction. So, we have $\left(\frac{3}{2}\right)^{(4n+2)-2}=5.0625^n<k_{4n+2}<5^n$, contradiction, so we're done.

Assume the contradiction, that $b \geq \frac{c+2a}{3}$ for all $a<b<c$. If we let $A$ be $a_1, a_2, \dots, a_{4n+2}$ in increasing order, this condition implies that \[ a_i \geq \frac{2a_{i-1} + a_{4n+2}}{3}\]for $i$ from $2$ to $4n+2$. In other words, each term is at least $\frac{1}{3}$ of the way between the previous term and the largest term. Let $b_i$ be $a_{4n+2} - a_{4n+2-i}$. We have that $b_{i+1} \geq \frac{3}{2} b_{i}$, hence $b_{i+4} \geq \frac{81}{16} b_{i} > 5b_{i}$. Thus, \[ b_{4n+1} > 5b_{4n-3} > \dots > 5^{n} b_{1}, \]contradiction.

Let $A = {a_1,a_2,\dots,a_{4n+2}}$ where $a_1<a_2<\dots<a_{4n+2}$. Fix $c = a_{4n+2} = max(A)$. Let $d_i = c - a_i$. Note that the condition is equivalent to $2d_a<3d_b$. BWOC, assume $2d_i \geq 3d_{i+1}$ for all $i$. Now, we see that, $$\frac{2}{3} d_{4n+1}\geq d_{4n+2}$$and $$\frac{2}{3} d_{4n}\geq d_{4n+1}$$and so on which gives us that $(\frac{2}{3})^{4n}d_1 \geq d_{4n+2}$, and optimizing we can have $max(d_1)=5^n$, even then this inequality implies $80^n \geq 81^n$ which is clearly false. Thus, there must exist some $i$ which satisfies the required conclusion. Thus, there exist $a,b,c \in A$ such that $a<b<c$ and $c+2a<3b$, as required.

My first ISL solve: We will prove this using contradiction. For the sake of contradiction, assume that the given statement is false. Thus, given a triple of three elements $a<b<c$ from the set $A$, we always have $c+2a \le 3b$. Suppose the subset of $A$ is $\{x_1, x_2, \dots, x_{4n+2}\}$ such that $x_i<x_j$ iff $i<j$. In particular, we must have $x_{4n+2}+2x_{k} \le 3x_{k+1}$ for all $k=1, 2, \dots, 4n$. We rewrite the equation as: \[3x_{4n+2}+2x_k \le 2x_{4n+2}+3x_{k+1}\]\[\implies 3(x_{4n+2}-x_{k+1}) \le 2(x_{4n+2}-x_k).\] Thus, we can simplify this equation to get: \[x_{4n+2}-x_k \ge \frac{3}{2}(x_{4n+2}-x_{k+1}).\] We apply this repeatedly to get: \[x_{4n+2}-x_k \ge \left (\frac{3}{2} \right)^{4n-k+1}(x_{4k+2}-x_{4k+1}).\] Now, we can plug in $k=1$: \[x_{4n+2}-x_1 \ge \left (\frac{3}{2} \right)^{4n}(x_{4k+2}-x_{4k+1}).\] From here, we look at both sides separately. Notice that $x_{4k+2}-x_1 \le 5^n$ since that is the maximum difference possible. Also, $x_{4k+2}-x_{4k+1} \ge 1.$ since that is the minimum difference possible. Finally, we can simplify $\left (\frac{3}{2} \right)^{4n}$ as $\left (\frac{81}{16} \right)^{n}>5^n$. Thus, the left hand side is less than the right hand side, a contradiction. Thus, there must be at least one triple $(a,b,c)$ such that $a<b<c$ and $c+2a>3b$. $\square$

Call a set of nonnegative integers good if given any three numbers, the middle one is at least 1/3 of the way from the small one to the large one. Let $f(n)$ denote the maximum possible number of elements other than 0 and $n$ of a good set with minimial element 0 and maximal element $n$. Suppose the smallest nonzero element in set is $k$. Then, by the condition, $k\geq n/3$. Then, between $k$ and $n$, we can get at most $f(n-k)$ elements beyond that. Clearly, $f$ is nondecreasing. If $n=3m$, then the best case is $k=m$, so $$f(3m)=1+f(2m).$$If $n=3m+1$, the best case is $k=m+1$, so $$f(3m+1)=1+f(2m).$$Finally, if $n=3m+2$, the best case is $k=m+2$, so $$f(3m+2)=1+f(2m).$$To summarize, $$f(n)=1+f(2\lfloor n/3 \rfloor).$$Define $f$ on all real numbers in an arbitrary manner as long as it preserves the fact that $f$ is nondecreasing. We have $$f(n)=1+f(2\lfloor n/3 \rfloor)\leq 1+f(2n/3),$$and with the base case $f(1)=0$ we have that $$f(n)\leq \log_{3/2}(n).$$Thus, $$f(5^n)+2\leq \log_{3/2}(5^n)+2<4n+2,$$hence done.

Assume otherwise. Then for any $a < b < c$ we must have $c + 2a \leq 3b$. Now let the subset of $A$ be denoted by $S$ with elements $(e_i)_{i=1}^{4n+2}$, where $e_i < e_j$ if and only if $i < j$. Fix $e_{4n+2}$ as $c$. Then we have $$2e_i + e_{4n+2} \leq 3e_{i+1}$$for all $0 < i < 4n+2$. Rewrite this as $$\frac{2}{3}e_i + \frac{1}{3}e_{4n+2} \leq e_{i+1}$$for all valid $i$. Clearly then we find, the inequality chain: \begin{align*} e_2 &\geq \frac{2}{3}e_1 + \frac{1}{3}e_{4n+2}\\ e_3 &\geq \frac{2}{3}e_2 + \frac{1}{3}e_{4n+2}\\ &\vdots\\ e_{4n+1} &\geq \frac{2}{3}e_{4n} + \frac{1}{3}e_{4n+2} \end{align*}Note that plugging in the first inequality into the second we find, \begin{align*} e_3 &\geq \frac{2}{3} \left(\frac{2}{3}e_{4n+2} + \frac{1}{3}e_{1} \right) + \frac{1}{3}e_{4n+2}\\ e_3 &\geq \frac 59 e_{4n+2} + \frac 49 e_1 \end{align*}Continually inducting we find, \begin{align*} e_{4n+1} \geq \frac{3^{4n} - 2^{4n}}{3^{4n}} e_{4n+2} + \frac{2^{4n}}{3^{4n}}e_1 \end{align*}Now note that, \begin{align*} \left( \frac{2}{3} \right)^{4n} (e_{4n+2} - e_1) \geq e_{4n+2} - e_{4n+1} \end{align*}Now noting that $e_{4n+2}-e_1 \leq 5^n$ we find, \begin{align*} \left(\frac{2}{3}\right)^{4n} \cdot 5^n > e_{4n+2} - e_{4n+1}\\ \left(\frac{80}{81}\right)^n > e_{4n+2} - e_{4n+1}\\ e_{4n+1} + 1 > e_{4n+2} \end{align*}However this is impossible.

Order the elements of $A$ as $a_1 < a_2 < \ldots < a_{4n+2} = M$. FTSOC, assume for all $i \in \{1, \ldots, 4n\}$, we have \[2a_i + M \leq 3a_{i+1}.\] Define $d_k = M - a_k$. Then our equation can be rewritten as \[3M - 3a_{i+1} \ge 2M - 2a_i \implies d_i \ge \frac 32 d_{i+1}.\] However, this tells us \[d_1 \ge \left(\frac 32\right)^{4n} \cdot d_{4n+1} = \left(\frac{81}{16}\right)^n \cdot d_{4n+1} > 5^n \cdot d_{4n+1},\] contradiction. Hence there exists some value $i = j$ such that our assumption was violated, giving us our desired triple $(a_j, a_{j+1}, a_{4n+2})$. $\blacksquare$

This is a really great problem! Assume for contradiction that no three elements with the given property exist. On the real number line, draw points $P_0$ and $Q$ at $0$ and $\max(A)$. Every minute, construct a new point $P_{i+1} = \tfrac{2}{3}P_i+\tfrac{1}{3}Q$, until we hit $k$ such that $P_kQ \ge 1$ and $P_{k+1}Q < 1$. Realize that $k+2$ is an upper bound on $|A|$. The key idea is that $P_iQ = (2/3)^i \cdot PQ \le (2/3)^i \cdot 5^n$. Thus, $(2/3)^k \cdot 5^n \ge 1$. This is achieved if and only if $k < 4n$, so $|A|<4n+2$, a contradiction.

Without loss of generality assume the elements of $A$ are such that $0 \leq a_0 < a_1 < a_2 \cdots < a_{4n + 1} \leq 5^n$. By way of contradiction, suppose that $3b \geq 2a + c$ for all triples $a, b, c \in A$. Then \begin{align*} 3a_k &\geq 2a_{k - 1} + a_{4n + 1}, \forall 1 \leq k \leq 4n \\ \implies 3a_k - 3a_{4n + 1} &\geq 2a_{k - 1} - 2a_{4n + 1} \\ \implies 3(a_{4n + 1} - a_k) &\leq 2(a_{4n + 1} - a_{k - 1}). \\ \end{align*}Let $b_k = a_{4n + 1} - a_k$. Then from the above we have that \[b_{k + 1} \leq \frac{2}{3}b_k \implies 1 \leq b_{4n} \leq b_0\left(\frac{2}{3}\right)^{4n} \leq 5^n \left(\frac{2}{3}\right)^{4n} = \left(\frac{80}{81}\right)^{n} < 1, \]which is a contradiction. $\blacksquare$

Let $A=\{x_1, x_2,\dots, x_{4n+2}\}$. The main idea is to look for triples of the form $(x_i, x_{i+1}, x_{4n+2})$. Suppose the contrary, i.e. for all $1\le i\le 4n+1$, we have $$3x_{i+1}\ge 2x_i+x_{4n+2}\iff 2(x_{i+1}-x_i)\ge x_{4n+2}-x_{i+1}$$ Let $y_i=x_{4n+2}-x_i$. Thus, $y_{i+1}=x_{4n+2}-x_{i+1}$ and $y_i-y_{i+1}=x_{i+1}-x_i$. This means that we can rewrite the inequality as $$2y_i-2y_{i+1}\ge y_{i+1}\iff 2y_i\ge3y_{i+1}\iff \frac{y_{i+1}}{y_i}\le \frac{2}{3}.$$Also, note that we have $y_{4n+1}=x_{4n+2}-x_{4n+1}\ge 1$ and $y_1=x_{4n+2}-x_1\le 5^n$ From here it follows that $$\left(\frac{1}{5}\right)^n\le\frac{y_{4n+1}}{y_1}\le \left(\frac{2}{3}\right)^{4n}=\left(\frac{16}{81}\right)^n\iff 81\le 80,$$which is a contradiction, so we are done.

FTSOC assume there is no $a<b<c$ in $A$ such that $c+2a>3b$. Then, $b-a\ge \frac{c-b}2$ for $a<b<c$. Let the elements of $A$ be $\{x_1, x_2, \dots, x_{4n+1}, x_{4n+2}\}$ such that $x_1>x_2>\dots>x_{4n+1}>x_{4n+2}$. Let $d=x_1-x_2\ge 1$. Then $x_2-x_3\ge \frac{d}{2}$ so $x_1-x_3\ge \frac{3d}{2}$. Then, $x_3-x_4\ge \frac{3d}{4}$, so $x_1-x_4\ge \frac{9d}{4}$. Therefore, $x_1-x_{4n+2}\ge(\frac32)^{4n}d\ge(\frac{81}{16})^nd\ge 5^n$, which is impossible, as desired for our contradiction.

Assume for the sake of contradiction that this doesn't hold. So for any positive integer $n=k \geq 1$, we have elements $x_0<x_1<...<x_{4k+1}$ such that $x_{4n}+2x_i \leq 3x_{i+1}$. Thus we find that $3(x_{4k+1}-x_{i+1}) \leq 2(x_{4k+1}-x_i) \implies \frac{3}{2} (x_{4k+1}-x_{i+1}) \leq x_{4k+1}-x_i$. Multiplying over all possible triples $(x_i, x_{i+1}, x_{4k+1})$ and dividing out common terms on both sides $x_{4k+1}-x_0 \geq (\frac{3}{2})^{4k}(x_{4k+1}-x_{4k}) = \frac{81}{16}^k (x_{4k+1}-x_{4k})$. Note that $\frac{81}{16}^k (x_{4k+1}-x_{4k}) \geq 5^k \geq x_{4k+1}-x_0$. Thus this is a contradiction and the problem statement holds.

Suppose the elements of A are $a_1 < a_2 < \cdots < a_{4n+2}$. Choose c = $a_{4n+2}$, a = $a_i$, b = $a_{i+1}$. If we prove that $c + 2a_i > 3a_{i+1}$ for at least one i, we are ready. Suppose the opposite $\Rightarrow$ $a_{i+1} \geq \frac{c+2a_i}{3}$, for all i. We set $a_i = b_i + c$ $\Rightarrow$ we get $b_{i+1} \geq \frac{2b_i}{3}$ $\Rightarrow$ $b_{4n+1} \geq \left(\frac{2}{3}\right)^{4n}b_1$ $\Rightarrow$ $\left(\frac{2}{3}\right)^{4n}(c-a_1) \geq c - a_{4n+1}$. However $\left(\frac{2}{3}\right)^{4n}(c-a_1) \leq \left(\frac{2}{3}\right)^{4n} \cdot 5^n = \left(\frac{80}{81}\right)^{n} < 1$. Also $c - a_{4n+1} \geq 1$ $\Rightarrow$ there is a contradiction $\Rightarrow$ we are ready.

Suppose that the elements of our subset are $x_1, x_2, x_3 \dots x_{4n+2}$ with $0 \leq x_1 < x_2 < x_3 \dots x_{4n} < x_{4n+1} < x_{4n+2} \leq 5^n$. Define $d_i = x_{i+1} - x_{i}$ for $1 \leq i \leq 4n+1$. Now assume for the sake of contradiction that there exist no $i, j, k$ such that $1 \leq i < j < k \leq 4n+2$ and $x_k + 2x_i > 3x_j$. So it follows that for all $i, j, k$ such that $1 \leq i < j < k \leq 4n+2$, we have $x_k + 2x_i \leq 3x_j \implies x_k - x_j \leq 2(x_j - x_i)$. So for $1 \leq a \leq 4n$, we have that $2(x_{a+1} - x_{a}) \geq x_{4n+2} - x_{a+1}$. It follows that $2d_{a} \geq d_{a+1} + \dots d_{4n+1}$ So it follows that $2d_{4n} \geq d_{4n+1}, 2d_{4n-1} \geq d_{4n} + d_{4n+1}, 2d_{4n-2} \geq d_{4n-1} + d_{4n} + d_{4n+1}, \dots, 2d_{1} \geq d_{2} + d_{3} \dots + d_{4n+1}$. So $d_{4n} \geq \frac{d_{4n+1}}{2}$. Next, $d_{4n-1} \geq \frac{ \frac{d_{4n+1}}{2} + d_{4n+1} }{2} = \frac{3d_{4n+1}}{4}, d_{4n-2} \geq \frac{ \frac{3d_{4n+1}}{4} + \frac{d_{4n+1}}{2} + d_{4n+1} }{2} = \frac{9d_{4n+1}}{8} $ for $k \in \{0, 1, 2, \dots 4n-1 \}$. We speculate that $d_{4n-k} \geq \frac{3^{k}}{2^{k+1}}d_{4n+1}$. This can be proven through strong induction. Base Case: Suppose $k = 0$, then it follows that $2d_{4n} \geq d_{4n+1} \implies d_{4n} \geq \frac{d_{4n+1}}{2}$ which is consistent with our conjecture. Inductive Step: Suppose that for some $0 \leq j \leq 4n-2$ we have that for all $0 \leq k \leq j$, $d_{4n-k} \geq \frac{3^{k}d_{4n+1}}{2^{k+1}}$. Now we know that $2d_{4n-(j+1)} \geq d_{4n-j} + d_{4n-(j-1)} + \dots d_{4n+1} $. So $d_{4n-(j+1)} \geq \frac{d_{4n-j} + d_{4n-(j-1)} + \dots d_{4n+1}}{2} \geq \frac{d_{4n+1} + \sum_{k=0}^{j}\frac{3^{k}d_{4n+1}}{2^{k+1}} }{2} \geq \frac{d_{4n+1}}{2} (1 + \frac{1}{2}\frac{(\frac{3}{2})^{j+1} - 1 }{\frac{3}{2} - 1} ) \geq d_{4n+1}\frac{3^{j+1}}{2^{j+2}} $. This completes the inductive step. Now we have that $d_1 + d_2 + d_3 \dots d_{4n+1} \leq 5^n$. But $d_1 + d_2 + d_3 \dots d_{4n+1} \geq d_{4n+1} + \sum_{k=0}^{4n-1}\frac{3^{k}}{2^{k+1}}d_{4n+1} = d_{4n+1}(1 + \frac{1}{2}\frac{(\frac{3}{2})^{4n}-1}{\frac{3}{2}-1}) = d_{4n+1}(\frac{3}{2})^{4n}$. Since $d_{4n+1} \geq 1$, it follows that $d_1 + d_2 + d_3 \dots d_{4n+1} \geq (\frac{3}{2})^{4n} = (\frac{81}{16})^n > 5^n$

Order the elements of $A$ as $a_1 \dots a_{4n + 2}$. If no such triples exist, we have $a_{4n + 2} \le 3a_{i + 1} - 2a_i$ for all $i$, or $3(a_{4n + 2} - a_{i + 1}) \le 2(a_{4n + 2} - a_i)$, so letting $b_i = a_{4n + 2} - a_i$, we have $3b_{i + 1} \le 2b_{i}$, so $b_1 \ge (\frac 32)^{4n} b_{4n + 1} \ge (\frac 32)^{4n} > 5^n$, so $a_{4n + 2} - a_{n} > 5^n$, but this is impossible since the maximal difference between any two elements of $A$ is $5^n$.

A nice problem

Let $A = \{ a_1 < a_2 < \cdots < a_{4n+2} \}$ and $a_{4n+2} - a_{i} = d_i$. Assume for contradiction that the given condition is never true. Applying the condition for contradiction on the triplet $(a_{i-1}, a_i, a_{4n+2})$, we have $d_{i} < \tfrac{2}{3} \cdot d_{i-1}$. Chaining inequalities, we have \begin{align*} 5^n \cdot \left( \frac{2}{3}\right)^{4n} \ge \left( \frac{2}{3}\right)^{4n} \cdot d_1 > \cdots > d_{4n+1}\end{align*}implying that $d_{4n+1} = a_{4n+2} - a_{4n+1} < (\tfrac{80}{81})^n < 1$, contradiction.

Wasted too much time on global stuff and chaining the wrong inequalities in this one.. Let the numbers be $a_1 < a_2 < \dots < a_{4n+2}$ and assume FTSOC $c+2a \leq 3b$. The entire idea of the problem is to write this as $\frac32 (c-b) \leq (c-a)$. Now we let $c=a_{4n+2}$ and exploit the confinment of $5^n$: \begin{align*} a_{4n+2} - a_1 &\geq \frac32 (a_{4n+2}-a_2) \\ &\geq \left( \frac32 \right)^2 (a_{4n+2}-a_3) \\ &\geq \ddots \\ &\geq \left( \frac32 \right)^{4n} (a_{4n+2} - a_{4n+1}) \end{align*}But using the miraculous fact that $\left( \frac32 \right)^{4n} = \left( \frac{81}{16} \right)^n > 5^n$ we get an obvious contradiction.