<s>[INST] 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!
 [/INST] Let's solve
$$ \min \Bigg[\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2} \Bigg]$$
subject to
$$
x+y+z=3 \\
x\ge0 \\
y\ge 0 \\
z\ge 0
$$
Formally, this is done with Kuhn-Tucker condition. Write those down, exploit symmetries, note that non-negativity constraints do not bind.
Solution: $x=y=z=1$. Hence, the minimum value of the objective is $6$.
The other inequality admits the same proof.
details: see wiki or original paper for details of KKT conditions.
$$
L(x,y,z,\lambda,\mu_1,\mu_2,\mu_3) =  -\Bigg[\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2} \Bigg] + \lambda[x+y+z=3] + \mu_1[x-0] +\mu_2[y-0] +\mu_3[z-0]
$$
KKT conditions:
\begin{align}
[x]\qquad
-\frac{1}{2}\eta_1 2x -\frac{1}{2}\eta_2 z-\frac{1}{2}\eta_3 y &= \lambda - \mu_1 \\
[y]\qquad
-\frac{1}{2}\eta_1 z -\frac{1}{2}\eta_2 2y-\frac{1}{2}\eta_3 x &= \lambda - \mu_2 \\
[z]\qquad
-\frac{1}{2}\eta_1 y -\frac{1}{2}\eta_2 x-\frac{1}{2}\eta_3 2z &= \lambda - \mu_3 \\
[\lambda]\qquad\qquad\qquad\qquad\quad
x+y+z&=3\\
[\mu_1]\qquad 
\mu_1\ge 0\qquad\text{and}\qquad \mu_1x&=0 \\
[\mu_2]\qquad
\mu_2\ge 0\qquad\text{and}\qquad \mu_2y&=0 \\
[\mu_3]\qquad
\mu_3\ge 0\qquad\text{and}\qquad \mu_3z&=0 \\
\end{align}
where $\eta_1 = (x^2+yz+2)^{-1/2},\eta_2 = (y^2+zx+2)^{-1/2}$ and $\eta_1 = (z^2+xy+2)^{-1/2}$
KKT methods states that the minimum of the objective should be a solution (for some $\lambda,\mu_1,\mu_2,\mu_3$) to the above system.
Wlog, let $x=\max\{x,y,z\}$ so that $x>0$ and $\mu_1=0$.
Case 1: $y=z=0$. Solve for $\lambda$ from $[x]$ and note that $\mu_2 <0$, therefore this case is empty.
Case 2: $y=0$ and $z=3-x\ne0$. Then $\mu_3=0$. Solve for $\lambda$ in $[x],[z]$, equate the two expressions and deduce that $x=z$. Observe that $\mu_2<0$ (at $x=z$), therefore this case is empty.
Case 3: $\mu_1=\mu_2=\mu_3=0$ so that non-negativity constraints are slack. Equate LHS of conditions $[x],[y],[z]$, combine with condition $[\lambda]$ to solve for $x,y,z$.
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