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For real numbers $x,y,z$ ,prove that $$x(x+y)^5+y(y+z)^5+z(z+x)^5 \geq \frac{32}{243}(x+y+z)^6$$

For $x,y,z>0$ I have a simple solution:

By Hölder inequality: $$\sum x(x+y)^5 \geq \frac{(\sum x(x+y))^5}{(\sum x)^4}\ge \frac{(\frac{2}{3}(\sum x)^2)^5}{(\sum x)^4}=\frac{32}{243}(x+y+z)^6 $$

But for $x,y,z \in \mathbb{R}$ the solution failed.

How can I prove the inequality for $x,y,z \in \mathbb{R}$?

lapcal
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  • @CalvinLin Do you mean $(\sum_{cyc} x(x+y)^5)(x + y + z)^4 \ge (\sum_{cyc} x(x+y))^5 $ does not hold for $x = -3, y = 1, z = 1$ or even $x = -2, y = 1, z= 1$? – River Li Feb 18 '22 at 02:33
  • @RiverLi Can you write the SOS expression of this inequality? – lapcal Feb 20 '22 at 06:22
  • My SOS expression is too complicated. Hope someone can give simple SOS. – River Li Feb 20 '22 at 06:30
  • @RiverLi It's nice as well if you express it to Matrix form. By the way, You can upload the code if your SOS expression is too long. – lapcal Feb 20 '22 at 10:04

2 Answers2

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Edit (2024-02-22). An SOS (Sum of Squares) solution simpler than my old one

Remarks: The proof is motivated by computer. It is still complicated to verify that $Q$ is positive semidefinite by hand. Hope to see a proof which can be verified by hand (the proof itself can be motivated by computer).

Let $a := x + y, b := y + z, c := z + x$. We have the following SOS expression $$x(x + y)^5 + y(y + z)^5 + z(z + x)^5 - \frac{32}{243}(x + y + z)^6 = \frac{1}{972}Z^\mathsf{T} Q Z$$ where $$Z := [{a}^{3},{a}^{2}b,a{b}^{2},{b}^{3},{a}^{2}c,abc,{b}^{2}c,a{c}^{2},b{c} ^{2},{c}^{3}], $$ and $$Q := \begin{pmatrix} A & B \\ B^\top & C \end{pmatrix} $$ with $$A := \left[ \begin {array}{ccccc} 484&-249&-217&-89&237 \\-249&404&69&-165&-38\\-217&69& 300&237&-150\\ -89&-165&237&484&-20 \\ 237&-38&-150&-20&300\end {array} \right], $$ and $$B := \left[ \begin {array}{ccccc} 8&100&-165&-20&-89\\ - 10&-40&-40&-31&100\\ 0&-38&-31&-150&-20 \\ 8&-249&100&-217&-89\\ 0&-31&69& -150&-217\end {array} \right] ,$$ and $$C := \left[ \begin {array}{ccccc} 6&-10&-10&0&8\\ -10& 404&-40&69&-165\\ -10&-40&404&-38&-249 \\ 0&69&-38&300&237\\ 8&-165&-249& 237&484\end {array} \right] .$$

River Li
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  • What’s the theory behind this? (Also there are some cubic terms?) – Maximilian Janisch Feb 23 '22 at 19:01
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    @MaximilianJanisch For basic idea of SOS, you may see Rodrigo de Azevedo's excellent answer https://math.stackexchange.com/questions/2410994/write-x2-y2-z22-3-x3-y-y3-z-z3-x-as-a-sum-of-three-squ/2414847#2414847 – River Li Feb 23 '22 at 23:35
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Let $$f(a,b,c)=a(a+b)^5+b(b+c)^5+c(c+a)^5- \frac{32}{243}(a+b+c)^6$$and the SOS tool gives the following result $$\begin{aligned} f(a,b,c) & = \frac{10}{818181}\sum \left(- 156 a^{3} - 544 a^{2} b + 143 a^{2} c - 403 a b^{2} + 395 a c^{2} + 149 b^{2} c + 260 b c^{2} + 156 c^{3}\right)^{2} \\ & + \frac{1}{4332331332}\left(\sum a \left(34418 a^{2} + 39597 a b - 14871 b^{2} - 59144 b c\right)\right)^{2} \\ & + \frac{1}{45175369079215116}\left(\sum a b \left(31282489 a - 450959 b - 30831530 c\right)\right)^{2} + \frac{48435918625}{218758445577}\left(\sum a b \left(b - c\right)\right)^{2} \end{aligned}$$ By the way, the original problem is true by Buffalo Way. See Wolfram Alpha

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