Set of numbers from $1-9$ multiplied together to get the smallest possible value The numbers $x_1,$ $x_2,$ $x_3,$ $y_1,$ $y_2,$ $y_3,$ $z_1,$ $z_2,$ $z_3$ are equal to the numbers $1,$ $2,$ $3,$ $\dots,$ $9$ in some order. Find the smallest possible value of
$$x_1 x_2 x_3 + y_1 y_2 y_3 + z_1 z_2 z_3.$$
I would assume the lowest number, $1,$ would have to be multiplied by $9,$ the highest.  I do not know how to approach this with AM-GM, though.
 A: The following Mathematica script confirms that there is only one solution:
prod[p_] := p[[1]] p[[2]] p[[3]] + p[[4]] p[[5]] p[[6]] + p[[7]] p[[8]] p[[9]];
perms = Permutations[{1, 2, 3, 4, 5, 6, 7, 8, 9}];
unique = Select[perms, (#[[1]] < #[[2]] < #[[3]] && #[[4]] < #[[5]] < #[[6]] && #[[7]] < #[[8]] < #[[9]] && #[[1]] < #[[4]] < #[[7]]) &];
products = Map[prod, unique];
min = Min[products];
result = Select[unique, (prod[#] == min) &];
Print[min, " ", result]

The script prints:
214 {{1, 8, 9, 2, 5, 7, 3, 4, 6}}

The next best two solutions are:
215 {{1, 7, 9, 2, 5, 8, 3, 4, 6}}
216 {{1, 8, 9, 2, 5, 6, 3, 4, 7}, {1, 8, 9, 2, 6, 7, 3, 4, 5}}

A: This isn't as systematic or as linear as I'd like but:
$\frac {M+N+P}3 \le \sqrt[3]{MNP}=\sqrt[3]{9!}\approx 71.3$ and equality is closest when the $w_1w_2w_3$ clusters are each closet to the geometric mean of $71.3$.
So we must have one term contaning the $9$.  That is, wolog, $9x_1x_2 \approx 71.3$ so $x_1x_2 \approx 7.9$.  Clearly the best option is $x_1x_2 = 8$ and $x_1,x_2 = 2,4$ or $1,8$.
$1,8$ has the advantage as it simultaneously finds clusters for the far off values of $1$ and $8$ as well.
In any event we also have to find, wolog, $7y_1y_2 \approx 71.3$ so $y_1x_2\approx 10.1$ and the best solution for that is $7,2,5$.
So with $9,1,8$ and $7,2,5$ that leaves $z_1, z_2, z_3 = 3,4,6$ and $3*4*6=72$ with our clusters being $70, 72, 72$ around our geometric mean of $71.3$ the only possible smaller sum would be be a tighter cluster and the only tighter cluster that is smaller it $71,71,71$, which is not an option.
So we can be assured that is the least.
We were "lucky" though
Still if say we had gotten something pretty close first, say,
$9*8*1 + 6*5*2 + 7*4*2 = 72 + 60 + 84$ we could see if we can improve it be attempting to finding a tighter cluster of  values closer to $71.3$, than $60$ and $84$ are.  The contenders that can formed by trios less than or equal to $9$ are:
$63=7*1*9; 64=8*2*4; 70=7*2*5; 72=9*8*1=9*2*4=6*3*4;80=8*2*5$ ... and it would take brute force to choose among those.  But even among that we want to get as close to $71.3$ as possible and we have $3$ ways to do $72$ and one way to do $70$.... well, that screams to uss to look if two of the ways to do $72$ are mutually compatible with the way to do $70$.
