Prove that, $$3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)≥abc(a+b+c)^3$$ For positive $a,b,c$

The exercises in this book are making me crazy. Any help would be appreciated.

My attempts:

I opened the LHS brackets and used the cube identity on RHS I get expressions which are somewhat similar but no idea how to proceed furthermore: $$3\sum_{cyc}(a^3b^3+a^4bc+a^2b^2c^2)\geq abc\sum_{cyc}(a^3+3a^2b+3a^2c+2abc).$$

  • 1
    $\begingroup$ I solved your problem. If you want to see my solution, show please your attempts. $\endgroup$ – Michael Rozenberg Feb 19 at 4:11
  • $\begingroup$ I opened the LHS brackets and used the cube identity on RHS, $\endgroup$ – Jordan Lawson Feb 19 at 4:15

By Holder $$3\prod_{cyc}a^2b\prod_{cyc}a^2c=\prod_{cyc}1\prod_{cyc}a^2b\prod_{cyc}a^2c\geq$$ $$\geq\left(\sum_{cyc}\sqrt[3]{1\cdot a^2b\cdot a^2c}\right)^3= \left(\sum_{cyc}\sqrt[3]{a^4bc}\right)^3=abc(a+b+c)^3.$$

  • $\begingroup$ Just got to know about Holder's inequality $\endgroup$ – Jordan Lawson Feb 19 at 4:25
  • $\begingroup$ Would have been better if I had known this earlier $\endgroup$ – Jordan Lawson Feb 19 at 4:26
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    $\begingroup$ Shouldn’t all the $\prod$ be replaced with $\sum$? $\endgroup$ – Mindlack Feb 19 at 14:52

Another way.

By your work, AM-GM and Muirhead we obtain: $$3\prod_{cyc}a^2b\prod_{cyc}a^2c-abc(a+b+c)^3=\frac{1}{2}\sum_{sym}(3a^3b^3+2a^4bc+a^2b^2c^2-6a^3b^2c)\geq$$ $$\geq \frac{1}{2}\sum_{sym}\left(6\sqrt[6]{(a^3b^3)^3(a^4bc)^2a^2b^2c^2}-6a^3b^2c\right)=3\sum_{sym}\left(a^{\frac{19}{6}}b^{\frac{13}{6}}c^{\frac{2}{3}}-a^3b^2c\right)\ge0.$$ The last inequality is true because $$\left(\frac{19}{6},\frac{13}{6},\frac{2}{3}\right)\succ(3,2,1).$$

  • $\begingroup$ What does that modified $>$ symbol mean? $\endgroup$ – preferred_anon Feb 19 at 12:30
  • $\begingroup$ $a\succ b$ says $a$ is a majoranta of $b$. Try to check it in google. I can not explain more because I am driving now. In evening, OK? $\endgroup$ – Michael Rozenberg Feb 19 at 13:05

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