From simple factorization: $a^9+b^9=(a+b)(a^2-ab+b^2)(a^6-a^3b^3+b^6)$. Since $(a-b)^2 \geq 0 \implies a^2-ab+b^2 \geq ab$. Similarly $(a^3-b^3) \geq 0 \implies a^6-a^3b^3+b^6 \geq a^3b^3$ So, $a^9+b^9 \geq (a+b)a^4b^4$. Similarly, $b^9+c^9 \geq (b+c)b^4c^4$ and $c^9+a^9 \geq (c+a)c^4a^4$ So, $(abc)^9=(a^9+b^9)(b^9+c^9)(c^9+a^9) \geq (a+b)(b+c)(c+a)a^8b^8c^8=a^9b^9c^9$ Equality is when $a=b=c$ or from first equation $8a^3=a^3$ which implies that at least one number of $a,b,c$ must be $0$