Sum of squares decomposition

Case $abc=0$. If, say, $c=0$, then the LHS is $a^2b^2\ge0$ and the RHS is $0$. Inequality holds.

Case $abc>0$. Divide by $abc$:
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c.$$

Substitution. Let $x=ab/c$, $y=bc/a$, $z=ca/b$. Then $xyz=abc$. Hmm, this is not obviously simpler.

Direct approach. By AM-GM, $a^2b^2+b^2c^2\ge2ab^2c$, etc. Summing gives $2(a^2b^2+b^2c^2+c^2a^2)\ge2abc(a+b+c)$. So $a^2b^2+b^2c^2+c^2a^2\ge abc(a+b+c)$.

Conclusion. The inequality holds, with equality iff $a=b=c$.

Remark. This is a nice AM-GM application: pair terms and bound by $2\cdot$ product.