Prove this inequality $3(abc+4)\ge 5(ab+bc+ca)$

I think your solution is true and very nice.

My proof.

We can assume that $ab+ac+bc\geq0$, otherwise the inequality is obvious.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove a linear inequality of $w^3$,

which says it's enough to prove our inequality for an extreme value of $w^3$,

which happens in the following cases.

1. $w^3=-4$.

In this case our inequality is obviously true;

2. $b=a$ and $c=3-2a$.

Hence, $$a^2(3-2a)\geq-4$$ or $$(a-2)(2a^2+a+2)\leq0,$$ which gives $a\leq2$ and we need to prove that $$3a^2(3-2a)+12\geq5(a^2+2a(3-2a))$$ or $$(a-1)^2(2-a)\geq0.$$ Done!