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 an explanation of Vasc's hint that you mentioned, which essentially solves the problem. Reproducing it here as:
$$\sum _{\text{cyc}} \sqrt{8\left(x^2+ \text{yz} + 2\right)}\geq  \sum _{\text{cyc}} \sqrt{(3x+y+z)^2+7} = \sum _{\text{cyc}} \sqrt{(2x+3)^2+7}\geq  12\sqrt{2}$$
Now the last inequality follows immediately from Minkowski's inequality and $\sum x=3$, so I will focus only on the first inequality below.
The inequality $\displaystyle \sum _{\text{cyc}} \sqrt{8\left(x^2+ \text{yz} + 2\right)}\geq \sum _{\text{cyc}} \sqrt{(2x+3)^2+7}$ can be established through Schur-concavity of the vector function $f{(\color{blue}{u}}) = \sqrt{u_1}+\sqrt{u_2}+\sqrt{u_3}$. This function is symmetric and concave, hence Schur-concave.  Alternately it is not difficult to establish the condition
$$\left(u_i- u_j\right)\left(\partial _{u_i}f-\partial _{u_j}f\right) = -\frac{\left(u_i- u_j\right){}^2}{\sqrt{u_i u_j}} \leq  0.$$  
<< If you are not familiar with Schur-concave property, it should be possible to derive similar results using the lemma $\sqrt{x-\epsilon }+\sqrt{y+\epsilon }\geq  \sqrt{x}+\sqrt{y}$.  This is easily shown to hold if $x-y \ge \epsilon \ge 0$ by squaring. Using the sequence $$[x, y, z] \succ [x - u, y + u, z] \succ [x - u, 
  y + u - (v + u), z + (u + v)]$$ and applying the lemma twice, one can establish $$\sqrt{x-u}+\sqrt{y-v}+\sqrt{z+u+v}\geq  \sqrt{x}+\sqrt{y}+\sqrt{z}$$ and the conditions needed.>>
As $f{(\color{blue}{u}})$ is Schur-concave, if we have $\color{blue}{a} \text{ and } \color{blue}{b}$ two vectors such that $\color{blue}{a}$ majorizes $\color{blue}{b}$, i.e. $\color{blue}{a} \succ  \color{blue}{b} \implies f(\color{blue}{a})\leq  f(\color{blue}{b})$.    In this case we need to show that considered as vectors,
$$\left[(2x+3)^2+7,(2y+3)^2+7, (2z+3)^2+7\right] \succ \left[8\left(x^2+\text{yz}+2\right),8\left(y^2+\text{zx}+2\right), 8\left(z^2+\text{xy}+2\right)\right]$$
If $s = xy+yz+zx$, we have
$$\sum _{\text{cyc}} \sqrt{8\left(x^2+\text{yz}+2\right)}= \sum _{\text{cyc}} \sqrt{8x^2+8\text{yz} + (x+y+z)^2+7} = \sum _{\text{cyc}} \sqrt{(3x+y+z)^2+7+ 4(2\text{yz}- \text{xy}-\text{zx})} = \sum _{\text{cyc}} \sqrt{(2x+3)^2+7+ 4(3\text{yz}-s)}$$
Let $\displaystyle a = (2x+3)^2+7, b = (2y+3)^2 + 7 \text{ and } c = (2z+3)^2+7$.  Also let $u = 4(s-3\text{yz}) \text{ and } v = 4(s-3\text{zx})$.  Now $\displaystyle x \geq  y \geq  z \Longrightarrow  \text{xy} \geq  \text{zx} \geq  \text{yz} \Longrightarrow  u \geq  0$ and $u+v \geq 0$ and the deviations are in reverse order to the ordering of $a, b, c$- hence giving the conditions needed for majorization.  It may be noted that in a crude sense, the deviations have mean $0$, and are ordered so as to bring the components "closer", hence increasing the concave function's value. 
As $[a, b, c] \succ  [a-u, b-v, c+(u+v)]$, we have $\displaystyle \sum _{\text{cyc}} \sqrt{8\left(x^2+ \text{yz} + 2\right)}\geq  \sum _{\text{cyc}} \sqrt{(2x+3)^2+7}$.  
Hope there is a simpler method to establish this.  Another possible approach could be to directly check if $\displaystyle F(x, y, z) = \sum _{\text{cyc}} \sqrt{x^2+ \text{yz} + 2}$ is Schur-convex, in which case any allowable $[x, y, z]\succ  [1, 1, 1]$ and the entire proof is done.  $F$ is symmetric, but we also need to check if $x-y)\left(\partial _xF -\partial _yF\right)\geq  0$ for this.
