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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

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

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  • $\begingroup$ 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
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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 $0<a<1.5.$

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  • $\begingroup$ 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

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