7
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
6
+25
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.