Mathematical Olympiad Rioplatense

a) The answer is yes, even under the weaker assumptions $a^b\le b^c$, $b^c\le c^d$, $c^d\le d^a$. The least number is $b$.

We need the inequality $\sqrt[n]{n}>\sqrt[n+1]{n+1}$ which holds for all $n\ge 3$. It is equivalent to $(1+\frac{1}{n}{)}^n<n$ and can be obtained by standard induction. The base $n=3$ is clear, and if $(1+\frac{1}{n}{)}^n<n$ for some $n$ then $${\left(1+\frac{1}{n+1}\right)}^{n+1}<{\left(1+\frac{1}{n}\right)}^n\left(1+\frac{1}{n+1}\right)<n\left(1+\frac{1}{n+1}\right)<n+1.$$ In particular $m>n\ge 3$ implies $\sqrt[n]{m}<\sqrt[n]{n}$. We use this general inequality to prove the following claim: If $u$, $v$, $w\in \mathbb{N}$ satisfy $u^v\le v^w$ then $w\ge u$ or $w\ge v$. Suppose on the contrary that $w<u$, $w<v$ and write $u^v\le v^w$ as $\sqrt[v]{v}\ge \sqrt[w]{u}$. Now $w<u$ implies $\sqrt[v]{v}\ge \sqrt[w]{u}>\sqrt[w]{w}$, which can hold only if $w\in \{1,2\}$. Indeed if $w\ge 3$ then $v>w\ge 3$, so the general inequality leads to the impossible $\sqrt[v]{v}<\sqrt[w]{w}$. For $w=2$ the condition is $u^v\le v^2$. Because $v>w=2$, this gives $v>u\ge 3$. Therefore $\sqrt[v]{v}<\sqrt[w]{u}$ by the general inequality. On the other hand $\sqrt[w]{u}<\sqrt[u]{2}$, so that $\sqrt[v]{v}<\sqrt[w]{u}<\sqrt[u]{u}$. However $\sqrt[v]{v}<\sqrt[u]{u}$ contradicts $u^v\le v^2$. Finally if $w=1$ then $u^v\le v$. On the other hand $u>1$, hence $u^v\ge 2^v>v$ for all $v\in \mathbb{N}$. The claim is proven. Given $a^b\le b^c$, $b^c\le c^d$, $c^d\le d^a$, we apply the claim to the triples $(a,b,c)$, $(b,c,d)$, $(c,d,a)$. Because $a$, $b$, $c$, $d$ are distinct, the conclusion is that none of $c$, $d$ and $a$ can be the least one among $a$, $b$, $c$, $d$. Therefore the least number is $b$.

b) The answer is no. Both quadruples $a=2^8$, $b=2$, $c=2^4$, $d=2^2$ and $a=2^8$, $b=2$, $c=2^4$, $d=2^9$ satisfy the condition. The greatest number in the first is $a=2^8$; the greatest number in the second is $d=2^9$.