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 a sketch of another alternative approach.
Substitute variables $x=a+1,y=b+1,z=c+1$; then $$f(a,b,c)=\sqrt{4+a^2+bc+a}+\sqrt{4+b^2+ca+b}+\sqrt{4+c^2+ab+c}$$ — I exploited the fact that $a+b+c=x+y+z-3=0$; also note, that $-1\le a,b,c \le 2$
Use Taylor formula $\sqrt{4+t}=2\sqrt{1+\frac{t}{4}}=2+\frac{t}{4}-\frac{t^2}{16}+R(t)$ to get Taylor expansion for $f$ up to second degree; I got $$64f=64\cdot6+16(a+b+c)+15(a^2+b^2+c^2)+16(ab+bc+ca)+R(a,b,c)=64\cdot6+7(a^2+b^2+c^2)+R(a,b,c)$$ — I again exploit that $(a+b+c)^2=0$
So, it is local minimum; after getting some bound for $R(a,b,c)$ one may get a value for $\varepsilon$ to conclude that for any $a,b,c$ that $|a|<\varepsilon,|b|<\varepsilon,|c|<\varepsilon$ we have $f(a,b,c)\ge 6$
Next, we remember that $a^2+bc+a=x^2+yz\ge0,...$, so $$|f'_a(a,b,c)|\le |\frac{2a+1}{2\sqrt{2}}+\frac{c}{2\sqrt{2}}+\frac{b}{2\sqrt{2}}|\le\frac{9}{2\sqrt{2}}$$ and so $|f'_b(a,b,c)|\le \frac{9}{2\sqrt{2}}$,  $|f'_c(a,b,c)|\le \frac{9}{2\sqrt{2}}$
Last step is just brute force calculation of values of $f$ in network of points $a=\frac{k}{N}$, $b=\frac{l}{N}$, $c=-\frac{k+l}{N}$ (except such values that lies in $|a|<\varepsilon,|b|<\varepsilon,|c|<\varepsilon$ where $f\approx 6$) where $N$ sufficiently large  and $-N\le (k,l,-k-l)\le 2N$; after such calculation one may observe that $f(\frac{k}{N}, \frac{l}{N}, -\frac{k+l}{N})>6+\frac{3\cdot 9}{2\sqrt{2}N}$ so there is no possibility that $f=6$, because $f(a,b,c)\ge f(\frac{k}{N}, \frac{l}{N}, -\frac{k+l}{N})-\frac{1}{N}\frac{9}{2\sqrt{2}}(|a-\frac{k}{N}|+|b-\frac{l}{N}|+|c-\frac{-k-l}{N}|)>6$
If current $N$ is not enough to observe that $f(\frac{k}{N}, \frac{l}{N}, -\frac{k+l}{N})>6+\frac{3\cdot 9}{2\sqrt{2}N}$, then one have to do brute force calculation with, say $N'=2N$; this will lead to success very soon, because $f''$ is small. I suspect that such calculation may be limited to <100 values, if one get good estimates for $R$.
