How prove this inequality $(a+b+c+d+e)^3\geq9(2abc+abd+abe+acd+ade+2bcd+bce+bde+2cde),$

Let $a,b,c,d$ and $,e$ are non-negatives .show that

$$(a+b+c+d+e)^3\geq9(2abc+abd+abe+acd+ade+2bcd+bce+bde+2cde),$$

Michael Rozenberg says that this inequality proof is very ugly.

I think this inequality seems nice and maybe it has simple methods to solve it?

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.

Finally, $$(a+b+c+d+e)^3 \ge 9((a+c)d(b+e) + (a+e)c(b+d) + (c+e)b(a+d)),$$ $$\boxed{(a+b+c+d+e)^3\geq9(2abc+abd+abe+acd+ade+2bcd+bce+bde+2cde).}$$