How to prove this inequality ??? 
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 

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.


*

*$w^3\rightarrow0$.


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 $0<a<1.5.$
