Prove this inequality $3(abc+4)\ge 5(ab+bc+ca)$ 
Let $\{a,b,c\}\subset\mathbb R$ such that $a+b+c=3$ and $abc\ge -4$. Prove that:  $$3(abc+4)\ge 5(ab+bc+ca).$$


*) $ab+bc+ca<0$ This ineq is right
*) $ab+bc+ca\ge 0$ then in $ab,bc, ca$ at least a non-negative number exists assume is $ab$
$\Rightarrow \displaystyle f(ab)=(3c-5)ab+5c^2-15c+12$
+)$\displaystyle 3c-5 > 0\Rightarrow \displaystyle f \geq 5c^2-15c+12=5(c-\frac{3}{2})^2+\frac{3}{4} > 0$
+) $\displaystyle 3c-5 \leq 0$. And we have: 
$\displaystyle \Rightarrow \frac{(3-c)^2}{4}+c(3-c) \geq ab+bc+ca \geq 0\displaystyle \Leftrightarrow -1 \leq c \leq \frac{5}{3}$
$\Rightarrow \displaystyle f \geq (3c-5)\frac{(3-c)^2}{4}+5c^2-15c+12 \geq 0$
$\displaystyle \Leftrightarrow (c-1)^2(c+1) \geq 0$
Help me to check up and post your solution. Thanks very much 
 A: 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. 


*

*$w^3=-4$.


In this case our inequality is obviously true;


*$b=a$ 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!
A: We have $a+ b+ c= 3$, in $a,\,b,\,c$ always at least a positive number exists. Assume $a> 0$, hypothesis
$$abc\geqq -\,4\,\therefore\,bc\geqq -\,\dfrac{4}{a}$$
On the other hand, we also have $t= bc\leqq \dfrac{(\!b+ c\!)^{\!2}}{4}= \dfrac{(\!3- a\!)^{\!2}}{4}$ and ${\rm A}\!=\!(\!3\,abc+ 4\!)- 5(\!ab+ bc+ ca\!)$
$$f(t)= 3at+ 12- 5t- 5a(\!3- a\!)= (\!3a- 5\!)t- 15a+ 5a^{2}+ 12$$ with $-\,\dfrac{4}{a}\leqq t\leqq \dfrac{(\!3- a\!)^{\,2}}{4}$. Consider two following cases


*

*$3\,a- 5= 0$ or $a= \dfrac{5}{3}$ then now
$$f(t)= -\,15\,.\,\frac{5}{3}+ 5\left ( \frac{5}{3} \right )^{\,2}+ 12- \frac{8}{9}> 0$$

*$3\,a- 5\neq 0$ then now $f(t)$ is a linear function for $t$ with $-\,\dfrac{4}{a}\leqq t\leqq \dfrac{(\!3- a\!)^{\,2}}{4}$, so to prove $f(t)\geqq 0$ is to prove $f\left ( -\,\dfrac{4}{a} \right )\geqq 0$ and $f\left ( \dfrac{(\!3- a\!)^{\,2}}{4} \right )\geqq 0$. Indeed, we have
$$f\left ( -\,\frac{4}{a} \right )= -\,\frac{4}{a}(\!3a- 5\!)- 15a+ 5a^{2}+ 12= \frac{1}{a}(\!5a^{3}- 15a^{2}+ 20\!)= \frac{5}{a}(\!a+ 1\!)(\!a- 2\!)^{2}\geqq 0$$
$$f(\!\dfrac{(\!3- a\!)^{\!2}}{4}\!)\!=\!\frac{(\!3a- 5\!)(\!3- a\!)^{\!2}}{4}\!-\!15a\!+\!5a^{2}\!+\!12\!=\!\frac{1}{4}(\!3a^{3}- 3a^{2}- 3a+ 3\!)\!=\!\frac{3}{4}(\!a+ 1\!)(\!a- 1\!)^{\!2}\!\geqq\!0$$
Therefore $f(t)\geqq 0$ with $-\dfrac{4}{a}\leqq t\leqq \dfrac{(\!3- a\!)^{\,2}}{4}$. Due to the same role of $a, b, c$ in ${\rm A}\geqq 0$, therefore 
$$\therefore\,3(\!abc+ 4\!)\geqq 5(\!ab+ bc+ ca\!)$$
