At first, from AM-GM:
$$\frac13(a+b+c+d+e)^3 \ge 9(a+b)c(d+e) = 9(acd+bcd+ace+bce),$$
and similarly for any permutations of $a,b,c,d,e.$
So it requres to find only 3 of them, the sum of which gives RHS of original inequality.
Taking in account that RHS of original inequality doesn'n contain the term $ace,$ one can obtain only 9 variants of satisfying permutations.
One of its combinations can lead to goal.
Easy to see that there are only 6 pairs of permutations which gives the term $2abc$ in sum. Third permutation can be found among another permutations.
Checking of them with Google Sheets gives required triple.
$$(a+b+c+d+e)^3 \ge 9((a+c)d(b+e) + (a+e)c(b+d) + (c+e)b(a+d)),$$