<s>[INST] Barnard and Child inequality exercise 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:
$$3\sum_{cyc}(a^3b^3+a^4bc+a^2b^2c^2)\geq abc\sum_{cyc}(a^3+3a^2b+3a^2c+2abc).$$
 [/INST] 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
$$\left(\frac{19}{6},\frac{13}{6},\frac{2}{3}\right)\succ(3,2,1).$$
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