Prove $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$, given $x+y+z=3$ and $x,y,z\ge0$ 
Let $x+y+z=3,x,y,z\ge 0$,show that
  $$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$$


Additional information 
I have seen the following problem:
$x,y,z>0,x+y+z=3$, prove that
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge 6.$$
Without loss of generality we can let $x=\max{\{x,y,z\}}$
Proof: case 1
$x\ge y\ge z$
we can easily prove 
$$\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge\sqrt{y^2+x+2}+\sqrt{z^2+z+2}$$
and
$$\sqrt{x^2+y+2}+\sqrt{y^2+x+2}\ge\sqrt{x^2+x+2}+\sqrt{y^2+y+2}$$
so we have
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge \sqrt{x^2+x+2}+\sqrt{y^2+y+2}+\sqrt{z^2+z+2}.$$
Then use
$$\sqrt{x^2+x+2}\ge\dfrac{3}{4}x+\dfrac{5}{4}$$
$$\sqrt{y^2+y+2}\ge\dfrac{3}{4}y+\dfrac{5}{4}$$
$$\sqrt{z^2+z+2}\ge\dfrac{3}{4}z+\dfrac{5}{4}$$
to get the result. Whereas the case 2 when $x\ge z\ge y$ can be proved using the same methods.

Now,I have another idea: using Holder inequality
we have
$$\left(\sum\sqrt{x^2+yz+2}\right)^2\left(\sum\dfrac{x^2+2yz+9}{x^2+yz+2}\right)\ge 36^3$$
$$\Longleftrightarrow \sum\dfrac{x^2+2yz+9}{x^2+yz+2}\le 1296$$
and  the following link has some discussion about this problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=538230
and 
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=538752&p=3097872#p3097872
and Vasc gave the hint:
$$\sum\sqrt{8(a^2+bc+2)}\ge \sum\sqrt{(3a+b+c)^2+7}\ge 12\sqrt 2$$
How prove this hint?Thank you everyone.
and my other idea is as follows:
let $a=\min(a,b,c)$
we can prove 
$$\sqrt{b^2+ca+2}+\sqrt{c^2+ab+2}\geq \sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\tag{1}$$
\begin{align*}
&\sqrt{a^2+bc+2}+\sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\\
\geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8} \tag{2}
\end{align*}
Summing up
\begin{align*}
&\sum_{cyc}{\sqrt{a^2+bc+2}}\\
\geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8}\\
=&\sqrt{a^2+\frac{(3-a)^2}{4}+2}+\sqrt{(3-a)^2+2a(3-a)+8}
\end{align*}
By the way: someone said $(1)$ is wrong? why? can anyone give an example? And hopefully someone can use this method to prove this inequality? Thank you very much!

Here is my sketch, mostly to convince that alternative approach (i.e. inequalities instead of search for minimum) really may be possible. At the other hand, useful inequalities have "differential" nature (that is, I think they may be most easily proven by differentiation, but there may be other ways to proof). 
Let us denote $a=x^2+yz, b=y^2+zx, c=z^2+xy$; then
$$a+b+c=x^2+yz+y^2+zx+z^2+xy=x^2+y^2+z^2+\frac{1}{2}((x+y+z)^2-(x^2+y^2+z^2))=$$
$$=x^2+y^2+z^2+4.5-\frac{1}{2}(x^2+y^2+z^2)\ge\frac{1}{2}3+4.5=6$$
Starting from reasonably looking inequality $\sqrt{u+2}+\sqrt{v+2}+\sqrt{w+2}\ge\sqrt{u+v+w+18}$ (where $u,v,w\ge 0$) we let $u=a,v=b,w=c$ and conclude that $$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge\sqrt{24}=\frac{6}{\sqrt{3/2}}$$
Starting from another inequality $f(u,v,w)\equiv\sqrt{u+2}+\sqrt{v+2}+\sqrt{w+2}\ge g(u,v,w)\equiv\sqrt{3(u+v+w)+18} $ (where $u,v,w\ge 0$ and $u\approx v\approx w$) we let $u=a,v=b,w=c$ and conclude that $$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge\sqrt{36}=6$$
Last inequality was made so that $f(2,2,2)=g(2,2,2)$ and $f'_u(2,2,2)=g'_u(2,2,2)$, $f'_v(2,2,2)=g'_v(2,2,2)$, $f'_w(2,2,2)=g'_w(2,2,2)$ (and yes, we need to analyze $f''$ vs. $g''$ too). Last inequality DOES NOT hold when, say, $u=7,v=0,w=0$ (5.83 vs. 6.25) so one probably need another similar inequality to cover, say, cases like $u\approx3v\approx3w$

Note, that it may be possible that I shall not convert the sketch into real proof.
I do not want to say that this alternative approach (that closely mimics differentiation) is the only possible. There may be possible other approaches, for example, based on the following lemma: 
Let $U\ge u \ge \delta\ge 0$, then $\sqrt{U}+\sqrt{u}\ge\sqrt{U+\delta}+\sqrt{u-\delta}$
Such approach may be much more elegant, but I still had no success with it.
