This is a question derived from another exercise problem for high school math contests. I hope this new question is also interesting.
If $a>0$, $b>0$, and $c>0$ with $abc=1$, prove that $$2(a^3+b^3+c^3)+3\ge 3(ab^2+bc^2+ca^2).$$
However, I only managed to use weighted AM-GM inequality $$ \frac{\frac{2}{3}a^3 + \frac{4}{3}b^3 + abc }{\frac{2}{3}+\frac{4}{3}+1} \ge a^{3/3}b^{5/3}c^{1/3}= \left(ab^2\right)^{2/3} $$ and its cyclic permutations $a\rightarrow b$, $b\rightarrow c$ and $c\rightarrow a$ to obtain $$ 2(a^3+b^3+c^3)+3\ge 3\left[ \left(ab^2\right)^{2/3}+ \left(bc^2\right)^{2/3}+\left(ca^2\right)^{2/3} \right]. $$ But I do not know how to prove the question without the extra exponent $2/3$. Any hints are appreciated.