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


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?

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    $\begingroup$ Most inequalities I've seen here are not so nice and do not not have simple methods to solve. $\endgroup$ – Han de Bruijn Mar 30 '17 at 18:01
  • $\begingroup$ For conext, the source of this question is math.stackexchange.com/q/2201085 (as I found in the edit history). It's too late now, but I would have suggested that you put you bounty directly over there, which I presume is the question you actually care about. It's much nicer than this, which does not seem nice at all $\endgroup$ – echinodermata Apr 1 '17 at 18:32
  • $\begingroup$ Maybe you could compute all partial derivatives to show that after some value $a+b+c+d+e=x$, $(a+b+c+d+e)^{3}$ is growing faster than messy RHS, then just check the finite cases directly, hopefully there are not many. $\endgroup$ – Morph Apr 5 '17 at 9:33
  • $\begingroup$ @echinodermata There are two different questions.Issue question has a nice form and routine way to solution, and this vice versa. $\endgroup$ – Yuri Negometyanov Apr 10 '17 at 12:08
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    $\begingroup$ @Morph It is much easier to apply the method of Lagrange multipliers to the original equation. But it's not so interesting. $\endgroup$ – Yuri Negometyanov Apr 10 '17 at 12:13

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. picture

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

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    $\begingroup$ My congratulations! $\endgroup$ – Michael Rozenberg Apr 8 '17 at 22:13
  • $\begingroup$ @MichaelRozenberg Thank you! At first I thought that the question was specially thought out. And only then I saw how it came out of the routine task with the help of "bw". $\endgroup$ – Yuri Negometyanov Apr 10 '17 at 11:58

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