For any real positive numbers $a, b, c$, prove that $3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq abc(a+b+c)^3$ My progress is that I applied Hölder’s for this,

$3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} \geq (a+b+c)^3$


whereas  $3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} =  (1+1+1)(a^2b+b^2c+c^2a)(\frac{b}{c} + \frac{c}{a} + \frac{a}{b})$

And I was lost here. Can anyone check or continue this proof?

Now, you can use Holder so:
$$(1+1+1)\left(a^2b+b^2c+c^2a\right)\left(\frac{b}{c} + \frac{c}{a} + \frac{a}{b}\right)=\sum_{cyc}1\sum_{cyc}a^2b\sum_{cyc}\frac{a}{b}\geq$$
$$\geq\left(\sum_{cyc}\sqrt[3]{1\cdot a^2b\cdot\frac{a}{b}}\right)^3=(a+b+c)^3.$$
Another way.
We need to prove that:
$$\sum_{cyc}(3a^3b^3+3a^4bc+3a^2b^2c^2)\geq\sum_{cyc}(a^4bc+3a^3b^2c+3a^3c^2b+2a^2b^2c^2)$$ or
$$\sum_{cyc}(3a^3b^3+2a^4bc-3a^3b^2c-3a^3c^2b+a^2b^2c^2)\geq0,$$ which is true by Schur:
$$3\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0$$ and Muirhead:
$$\sum_{cyc}(2a^4bc-2a^2b^2c^2)\geq0.$$
