Prove that inequality $27+7\left(ab+bc+ca\right)\le 8\left(\sum \sqrt{a+3b}\right)$ For the positive reals $a,b$ and $c$ so that $a+b+c=3$. Show that $$27+7\left(ab+bc+ca\right)\le 8\left(\sqrt{a+3b}+\sqrt{b+3c}+\sqrt{c+3a}\right)$$

That inequality has been created by Imad Zak, i think it is interesting problem.
We can see that $$\text{L.H.S}=3(a+b+c)^2+7(ab+bc+ca)$$
$$=3(a^2+b^2+c^2)+13(ab+bc+ca)$$
$$=\sum_{cyc} (a+3b)(b+3c)$$
Let $\sqrt{a+3b}=x;\sqrt{b+3c}=y$ and $z=\sqrt{c+3a}$ where $x,y,z>0$ and $x^2+y^2+z^2=12$
We prove $$8(x+y+z)\ge x^2y^2+y^2z^2+z^2x^2$$
I am stuck here. I tried to homogenize the last inequality but i only get wrong inequalities. I also used C-S or Holder but failed.  
 A: Let $a^2+b^2+c^2=k(ab+ac+bc).$
Thus, by Holder
$$\sum_{cyc}\sqrt{a+3b}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{a+3b)}\right)^2\sum\limits_{cyc}(a+3b)^2}{\sum\limits_{cyc}(a+3b)^2}}\geq$$
$$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(a+3b)\right)^3}{\sum\limits_{cyc}(10a^2+6ab)}}=\sqrt{\frac{32(a+b+c)^6}{27\sum\limits_{cyc}(5a^2+3ab)}}.$$
Id est, it's enough to prove that
$$8\sqrt{\frac{32(a+b+c)^6}{27\sum\limits_{cyc}(5a^2+3ab)}}\geq3(a+b+c)^2+7(ab+ac+bc)$$ or
$$8\sqrt{\frac{32\left(\sum\limits_{cyc}(a^2+2ab)\right)^3}{27\sum\limits_{cyc}(5a^2+3ab)}}\geq\sum_{cyc}(3a^2+13ab)$$ or
$$8\sqrt{\frac{32(k+2)^3}{27(5k+3)}}\geq3k+13$$ or
$$2048(k+2)^3\geq27(5k+3)(3k+13)^2,$$ which is true by AM-GM:
$$27(5k+3)(3k+13)^2=\frac{27}{2}(10k+6)(3k+13)^2\leq$$
$$\leq\frac{27}{2}\left(\frac{10k+6+2(3k+13)}{3}\right)^3=2048(k+2)^3$$
and we are done!
A: You are actually very close. We can go on from your last inequality, with $x=2u, y=2v, z=2w$, we need to show
$$u+v+w \ge u^2v^2+v^2w^2+ w^2u^2\tag{1}$$
for $u^2+v^2+w^2=3$.
Now, the RHS of (1) is
$$\frac{(u^2+v^2+w^2)^2 - (u^4+v^4+w^4)}2 = \frac{9- (u^4+v^4+w^4)}2$$
So we need to prove
$$u^4+ v^4 + w^4 + 2(u+v+w) \ge9.$$
But this is clear since:
$$u^4 + u + u \ge 3\sqrt[3]{u^4\cdot u\cdot u} = 3u^2,$$
and so on. QED. 
