Notice that one can scale each $a_i$ since the sum of exponents of both sides of both inequalities are the same. Thus, wlog let $a_n=1$ where $a_1 \leq a_2 \leq \cdots \leq a_n$, we now have \[ a_2a_3 \cdots a_n \leq a_1 \] We will prove this inequality by induction on $n$, for $n=2$ we have $a_2 \geq a_1$ which is trivial. Now assuming $k-1$ works, we will show $k$ works. Indeed, since $a_{i-1} \leq a_i \leq a_k=1$ we have $\sqrt{a_{i-1}} \leq a_i$ so, to prove $a_1 \leq a_2a_3 \cdots a_k$ it is sufficient to show $a_1 \leq \sqrt{a_1a_2 \cdots a_{k-1}} \Leftrightarrow a_1 \leq a_2a_3 \cdots a_{k-1}$. Completing the induction. Now for equality. Given that $a_2a_3 \cdots a_n = a_1$ and $a_2a_3 \cdots a_n \geq \sqrt{a_1a_2 \cdots a_{n-1}} \geq a_1$ we must have $\sqrt{a_1a_2 \cdots a_{n-1}}=a_1 \Rightarrow a_2a_3 \cdots a_{n-1}=a_1$. Implying $a_n=1$. Inducting downwards convers all $a_i$. Scaling yields $a_1=a_2= \cdots =a_n=c$ for any positive real $c$. $\blacksquare$

Assume wlog $m=a_1\leq ...\leq a_n=M$. Note that for all $2\leq i\leq n-1$ we must have $a_i\geq \sqrt{a_na_{i-1}}=\sqrt{Ma_{i-1}}$. Now let $P=a_1\cdots a_n$. By our lemma we must have $$P=a_1\cdots a_n\geq a_1a_n\sqrt{Ma_1}\sqrt{Ma_2}\cdots \sqrt{Ma_{n-2}}=\sqrt{M^{n-1}m^2P/a_{n-1}}\implies$$$$P\geq \frac{M}{a_{n-1}}M^{n-2}m^2\geq M^{n-2}m^2$$For equality to happen we must have $a_{i-1}M=a_i^2\forall i=2,...,n-1$ and $M=a_{n-1}$. By downward induction starting from the base case $k=n-1$ it is easy to see that this only holds when $a_k=M$ for all $k$, i.e. all numbers are equal

Here is a unnecessarily complicated solution that I found when I compiled the shortlist for the competition. (I wanted to post this a long time ago but forgot about it). Hopefully it is instructive for someone. Let $b_i=\log{a_i}$ and WLOG let $b_1\leq b_2\dots \leq b_n$. This lets us change the product on the inequality for a sum. We define the function \[f(x)=b_n-\frac{b_n-b_1}{2^{x-1}}\]We graph $f(x)$ and also plot the points $(i,b_i)$ for all $1\leq i\leq n$. The condition of the problem implies that \[\frac{b_{i}+b_n}{2}\leq b_{i+1}\] We also know that \[\frac{f(i)+f(n)}{2}= f(i+1)\] This combined with the fact that $f(1)=b_1$ implies (by a straightforward induction) that $b_i\geq f(i)$. Which means that the points $(i,b_i)$ lie all above the graph of $f(x)$ and to minimize them we have to "drop" them onto the curve [asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.9647805568614999, xmax = 14.756076367985, ymin = -0.08243429514463951, ymax = 9.503454073664201; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); /* draw grid of horizontal/vertical lines */ pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 1, gridy = 1; /* grid intervals */ for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx) draw((i,ymin)--(i,ymax), gridstyle); for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy) draw((xmin,i)--(xmax,i), gridstyle); /* end grid */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ real f1 (real x) {return 7-5/2^(x-1);} draw(graph(f1,-0.9547805568614999,14.746076367985001), linewidth(2) + qqwuqq); /* dots and labels */ label("$f$", (0.6374322133536922,-0.013963663938862083), NE * labelscalefactor,qqwuqq); dot((1,2),dotstyle); label("$b_1$", (1.0482560005883568,2.1360141559225494), NE * labelscalefactor); dot((2,4.5),dotstyle); dot((3,5.75),dotstyle); dot((4,6.375),dotstyle); dot((5.02,6.691802279845825),dotstyle); dot((6,6.84375),dotstyle); dot((7,6.921875),dotstyle); dot((1.9913139637845407,5.158293946096148),dotstyle); label("$b_2$", (2.0479272161927073,5.299357317629467), NE * labelscalefactor); dot((3.007814747009537,6.460128282507111),dotstyle); label("$b_3$", (3.0612925580382133,6.6002993105392385), NE * labelscalefactor); dot((4,7),dotstyle); label("$b_4$", (4.060963773642564,7.134370233944302), NE * labelscalefactor); dot((4.987316272237161,7.1556288183979),dotstyle); label("$b_5$", (5.046940863005759,7.298699748838168), NE * labelscalefactor); dot((6.039483749610403,7.244795553768514),dotstyle); label("$b_6$", (6.087694457333576,7.380864506285101), NE * labelscalefactor); dot((7.002484491613031,7.280462247916759),dotstyle); label("$b_7$", (7.059977420455615,7.421946885008567), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] If we do the calculations we get \[b_1+b_2+\dots+b_n\geq f(1)+f(2)+\dots +f(n)=(n-2)b_n+2b_1+(b_n-b_1)(\frac{1}{2^n})\geq (n-2)b_n+2b_1 \]Equality happens when $b_n=b_1$. So we get that for equality to occur all the $b_i$ have to be equal.