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


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;

  1. $b=a$ 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!

  • 2
    $\begingroup$ (+1), when i saw no one answered the question within the first 5 min, i knew it would be you how will answer it if i get back to the question, you man like the master of these kind of problems, very nice solution. $\endgroup$ – Ahmad Sep 21 '17 at 17:39
  • $\begingroup$ Where can I find uvw theorem? $\endgroup$ – Word Shallow Sep 23 '17 at 10:49
  • $\begingroup$ @Word Shallow If you know Russian then there is my paper about it. Also see here: math.stackexchange.com/tags/uvw/info and here: artofproblemsolving.com/community/c6h278791 $\endgroup$ – Michael Rozenberg Sep 23 '17 at 10:52

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