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Let $a$, $b$ and $c$ be real numbers. Prove that: $$4(a^6+b^6+c^6)+5(a^5b+b^5c+c^5a)\geq\frac{(a+b+c)^6}{27}$$ I tried SOS, uvw and more, but without success.

For example.

Let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v^2$ may be negative, and $abc=w^3$.

Thus, we need to prove $$4(729u^6-1458u^4v^2+729u^2v^4-54v^6+162u^3w^3-108uv^2w^3+3w^6)+$$ $$+5\sum_{cyc}a^5b\geq27u^6$$ or $$8(729u^6-1458u^4v^2+729u^2v^4-54v^6+162u^3w^3-108uv^2w^3+3w^6)+$$ $$+5\sum_{cyc}(a^5b+a^5c)\geq54u^6+5\sum_{cyc}(a^5c-a^5b)$$ or $$9(642u^6-1161u^4v^2+468u^2v^4-18v^6+129u^3w^3-61uv^2w^3+w^6)\geq$$ $$\geq5(a-b)(b-c)(c-a)\sum_{cyc}\left(a^3+a^2b+a^2c+\frac{1}{3}abc\right)$$ or $$9(642u^6-1161u^4v^2+468u^2v^4-18v^6+129u^3w^3-61uv^2w^3+w^6)\geq$$ $$\geq5(a-b)(b-c)(c-a)\sum_{cyc}(27u^3-18uv^2+w^3).$$ We'll prove that $$642u^6-1161u^4v^2+468u^2v^4-18v^6+129u^3w^3-61uv^2w^3+w^6\geq0,$$ for which it's enough to prove $$642u^6-1161u^4v^2+468u^2v^4-18v^6+129u^3w^3-61uv^2w^3+w^6\geq(uv^2-w^3)^2,$$ which is a linear inequality of $w^3$.

Thus, by uvw it's enough to check $b=c=1$, which gives $$(a-1)^2(107a^4+337a^3+501a^2+529a+281)\geq0,$$ which is true for any real value of $a$.

Id est, it's enough to prove $$81(642u^6-1161u^4v^2+468u^2v^4-18v^6+129u^3w^3-61uv^2w^3+w^6)^2\geq$$ $$\geq25\cdot27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)(27u^3-18uv^2+w^3)^2,$$ which is fourth degree of $w^3$ and I'm at a dead end here.

Thank you, for your help!

Michael Rozenberg
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3 Answers3

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The following proof is true if they are non-negative reals.

$4(x^3+y^3+z^3)+15xyz\geq (x+y+z)^3$- this directly follows from Schur. So $$4(a^6+b^6+b^6)+5(a^5b+b^5c+c^5a)\geq 4(a^6+b^6+c^6)+15a^2b^2c^2\geq (a^2+b^2+c^2)^3\geq\dfrac{(a+b+c)^6}{27}$$

dezdichado
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With computer, here is a solution.

We have $$4(a^6+b^6+c^6) + 5(a^5b+b^5c+c^5a) - \frac{(a+b+c)^6}{27} = \frac{1}{324}z^TQz$$ where \begin{align} z = \left(\begin{array}{l} a^2\, b - a^3\\ a^2\, c - a^3\\ a^3 - 2\, a^2\, b + a\, b^2\\ a^3 - a^2\, c - a^2\, b + a\, b\, c\\ a^3 - 2\, a^2\, c + a\, c^2\\ - a^3 + 3\, a^2\, b - 3\, a\, b^2 + b^3\\ - a^3 + 2\, a^2\, b + c\, a^2 - a\, b^2 - 2\, c\, a\, b + c\, b^2\\ - a^3 + 2\, a^2\, c + b\, a^2 - a\, c^2 - 2\, b\, a\, c + b\, c^2\\ - a^3 + 3\, a^2\, c - 3\, a\, c^2 + c^3 \end{array}\right) \end{align} and \begin{align} Q = \left(\begin{array}{rrrrrrrrr} 21060 & -10530 & 17820 & 4515 & -8817 & 3685 & 4023 & -1722 & -1575\\ -10530 & 21060 & -6135 & -903 & 17820 & -882 & -2166 & 1785 & 3783\\ 17820 & -6135 & 18550 & 1719 & -4026 & 4590 & 3570 & -2286 & -285\\ 4515 & -903 & 1719 & 6108 & -3450 & -60 & 1485 & 1320 & -1332\\ -8817 & 17820 & -4026 & -3450 & 18354 & -279 & -2115 & 792 & 4590\\ 3685 & -882 & 4590 & -60 & -279 & 1284 & 774 & -642 & 126\\ 4023 & -2166 & 3570 & 1485 & -2115 & 774 & 1104 & -246 & -485\\ -1722 & 1785 & -2286 & 1320 & 792 & -642 & -246 & 790 & -36\\ -1575 & 3783 & -285 & -1332 & 4590 & 126 & -485 & -36 & 1284 \end{array}\right). \end{align} It is easy to verify that $Q$ is (real symmetric) positive definite. We are done.

River Li
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  • It's very hard! – NKellira Aug 08 '20 at 03:34
  • @knvy144444 Q is $9\times 9$ which is small. In some problems, $Q$ may be large which makes it difficult to find SOS. – River Li Aug 08 '20 at 03:36
  • I agree. But how do you get Q in Maple? – NKellira Aug 08 '20 at 03:37
  • I did not use maple to get SOS. I used matlab. – River Li Aug 08 '20 at 03:40
  • Wow, so what tool you use in Matlab? – NKellira Aug 08 '20 at 03:41
  • See my comments: https://mathoverflow.net/questions/239243/an-inequality-concerning-lagranges-identity/335440#335440 – River Li Aug 08 '20 at 03:56
  • @lsxpqr See my last comment. – River Li Feb 20 '22 at 05:18
  • Why does your $z$ look like that? Normally we would use something like $$z=\begin{pmatrix}x^3\x^2y\x^2z\xy^2\xyz\xz^2\y^3\y^2z\yz^2\z^3\end{pmatrix},$$ wouldn't we? – youthdoo Aug 15 '25 at 09:41
  • @youthdoo I used some substitution before doing SOS. – River Li Aug 15 '25 at 09:44
  • @youthdoo Anyway, now there are some solvers to give SOS. I think math110's answer uses such a solver. Will you post your solution using some solver? – River Li Aug 15 '25 at 09:52
  • @youthdoo Sometimes, after doing some substitution, SOSTools can output a real SOS, but without doing the substitution, SOSTools does not output a real SOS. This is why I use substitution. I don't consider the benefit. – River Li Aug 15 '25 at 10:05
  • I think I understand the benefit of your substitution. Please if the following makes sense – youthdoo Aug 15 '25 at 11:53
  • If we use the $z$ in my comment above, then for $a=b=c$ we have $z^TQz=0$ but $z\ne\bf0$, which means that $Q$ can be at most positive semi definite but not strictly positive definite. For the $z$ you used, we have $z=\bf0$ for $a=b=c$ (as well as $z^TQz=0$) which means that $Q$ might be strictly positive definite. If we find a specialization of the parametrized matrix $Q$ which makes it strictly positive definite, we can round it to a rational matrix (which is nicer) but we cannot do the rounding if $Q$ is only positive semi definite (or else it will produce small negative eig values) – youthdoo Aug 15 '25 at 11:54
  • @youthdoo If SOSTools outs a real PD Q, then it can output a rational Q as well. – River Li Aug 15 '25 at 12:23
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if we let $$f(a,b,c)=108\sum a^6+135 \sum a^5b-(a+b+c)^6$$ then we have

$$\begin{aligned} f(a,b,c) & = \frac{5}{373958}\sum \left(- 1514 a^{3} - 1597 a^{2} b - 757 a^{2} c + 1567 a c^{2} + 30 b^{2} c + 757 b c^{2} + 1514 c^{3}\right)^{2} \\ & + \frac{1}{1989865763207556}\left(\sum a b \left(- 178406783 a + 102251848 b + 76154935 c\right)\right)^{2} \\ & + \frac{1}{11153532}\left(\sum a \left(22578 a^{2} + 223 a b - 16622 b^{2} - 6179 b c\right)\right)^{2} \\ & + \frac{173777280}{178406783}\left(\sum a b \left(b - c\right)\right)^{2} \end{aligned}$$

math110
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