Let $a,b,c$ are positive real numbers such that $a+b+c=3$. Prove that $$\sum\frac{(7a^{3}+3)(b+c)}{7a+3} \geq 6 $$

I try to prove $LHS \geq \sum\frac{9}{5}a+\frac{1}{5}$ but don't succeed


closed as off-topic by Carl Mummert, Claude Leibovici, Taroccoesbrocco, user99914, Tyrone Jul 29 '18 at 12:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Claude Leibovici, Taroccoesbrocco, Community, Tyrone
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What is the sum across? Is it a cyclic sum? $\endgroup$ – jgon Jul 21 '18 at 5:54
  • $\begingroup$ yes, it is cyclic $\endgroup$ – Truth Jul 21 '18 at 6:03
  • $\begingroup$ @Unruly Kid Try the Vasc's LCF Theorem. Also, the $uvw$ helps. $\endgroup$ – Michael Rozenberg Jul 21 '18 at 6:27
  • $\begingroup$ @MichaelRozenberg Please post your full solution $\endgroup$ – Truth Jul 21 '18 at 6:31
  • $\begingroup$ @Unruly Kid By LCF or by $uvw$? $\endgroup$ – Michael Rozenberg Jul 21 '18 at 6:34

Yes, you are right, the TL method does not help here.

But $uvw$ helps.

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

Hence, we need to prove that $$\sum_{cyc}(7a^3+3u^3)(3u-a)(7b+3u)(7c+3u)\geq6u^3\prod_{cyc}(7a+3u)$$ and we see that our inequality it's $f(w^3)\geq0,$ where $$f(w^3)=-343\cdot3w^6+A(u,v^2)w^3+B(u,v^2).$$ We see that $f$ is a concave function, which says that it's enough to prove our inequality for an extreme value of $w^3$, which happens in the following cases.

  1. $w^3\rightarrow0$.

Let $c\rightarrow0$ and $b=3-a$, where $0<a<3$.

We obtain: $$(3-a)^2a^2\geq0;$$ 2. Two variables are equal.

Let $b=a$ and $c=3-2a$, where $0<a<1.5.$

We obtain: $$a^2(a-1)^2(39+70a-49a^2)\geq0.$$ Done!

A proof by LCF.

Let $f(x)=\frac{(7x^3+3)(x-3)}{7x+3}.$$

Hence, $$f''(x)=\frac{42(x+1)(49x^3-42x^2-3x-24)}{(7x+3)^3}<0$$ for all $0<x<1$ and we need to prove that $$\frac{f(a)+f(b)+f(c)}{3}\leq f\left(\frac{a+b+c}{3}\right).$$ Thus, by the Vasc's LCF Theorem it's enough to prove the last inequality for $b=a\leq1$ and $c=3-2a\geq1.$

After these substitutions we obtain $$a^2(3-2a)(39+70a-49a^2)\geq0,$$ which is true even for all $0<a<1.5.$

  • $\begingroup$ Thank you, and how to prove with LCF? $\endgroup$ – Truth Jul 21 '18 at 8:03
  • $\begingroup$ @Unruly Kid I added something. See now. $\endgroup$ – Michael Rozenberg Jul 21 '18 at 8:36
  • $\begingroup$ Nice solution. Thank you $\endgroup$ – Truth Jul 21 '18 at 8:39
  • $\begingroup$ @Unruly Kid You are welcome! I see also three solutions, but they are very ugly. $\endgroup$ – Michael Rozenberg Jul 21 '18 at 8:39
  • $\begingroup$ Why someone down voted? Explain please your step. $\endgroup$ – Michael Rozenberg Jul 25 '18 at 12:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.