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.

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.