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!
 A: 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 $x,y,z$.
A: 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.
A: 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.
A: 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 $R$.
