RMO Inequalities - 2

By $AM-GM$ inequality ,

$\large{\dfrac{a^2b+b^2c+c^2a}{3} \geq \sqrt[3]{(a^2b)(b^2c)(c^2a)} = \sqrt[3]{a^3b^3c^3} \Rightarrow a^2b+b^2c+c^2a \geq 3abc \dots (1) \\ \dfrac{ab^2+bc^2+ca^2}{3} \geq \sqrt[3]{(ab^2)(bc^2)(ca^2)} = \sqrt[3]{a^3b^3c^3} \Rightarrow ab^2+bc^2+ca^2 \geq 3abc \dots (2) \\ (1) \times (2) \Rightarrow (a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq 9a^2b^2c^2 \\ \Rightarrow \dfrac{(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)}{a^2b^2c^2} \geq \boxed{9}}$

Just to be different, I have a proof through Cauchy Schwarz.

$\begin{aligned} \dfrac {(a^2 b + b^2 c + c^2 a) (ab^2 + bc^2 + ca^2)}{a^2 b^2 c^2} &= \left ( \dfrac{a^2 b + b^2 c + c^2 a}{abc} \right ) \left ( \dfrac{a b^2 + b c^2 + c a^2}{abc} \right )\\ &= \left (\displaystyle \sum_{cyc} \dfrac{a}{b} \right ) \left (\displaystyle \sum_{cyc} \dfrac {b}{a} \right )\\ & \geq (1+1+1)^2\\ &= 9 \end{aligned}$

Therefore, the minimum value of the expression is $9$ .

Applying $AM-GM$ on $\left(\dfrac ac,\dfrac ba, \dfrac cb\right)$ : $\dfrac{\dfrac ac+\dfrac ba+ \dfrac cb}3\geq\sqrt[3]{\dfrac ac×\dfrac ba×\dfrac cb}\\a^2b+b^2c+c^2a\geq3abc$

Similarly, $ab^2+bc^2+ca^2\geq3abc\\\implies(a^2b+b^2c+c^a)(ab^2+bc^2+ca^2)\geq9a^2b^2c^2\\\dfrac{(a^2b+b^2c+c^a)(ab^2+bc^2+ca^2)}{a^2b^2c^2}\geq9$

Use AM -HM inequality. First divide the firs expression of the numerator with the denominator.Then use AM-HM inequality.