Inequality. $\frac{1}{\sqrt{x^2+yz+3}}+\frac{1}{\sqrt{y^2+zx+3}}+\frac{1}{\sqrt{z^2+xy+3}} \geq 1$ Prove that : 
$$\frac{1}{\sqrt{x^2+yz+3}}+\frac{1}{\sqrt{y^2+zx+3}}+\frac{1}{\sqrt{z^2+xy+3}} \geq 1$$ if $x^2+y^2+z^2 \leq9$. 
I try to apply Cauchy-Buniakowski and I obtaine the followin: 
$$\sum_{x,y,z}{\frac{1}{\sqrt{x^2+yz+3}}}\cdot \sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}\geq 9$$
So I have to prove that : $$\displaystyle\frac{9}{\sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}} \geq 1$$ if $x^2+y^2+z^2 \leq9$. 
Another trying :
$$\left(\sum_{x,y,z}{\sqrt{x^2+yz+3}}\right) \leq \sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)} $$ so we have to prove that: 
$$\frac{9}{\sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)}} \geq 1$$ hence:
$$3(x^2+y^2+z^2+xy+yz+zx+9) \leq 81$$ or
$$(x^2+y^2+z^2+xy+yz+zx+9) \leq 27$$ or
$$x^2+y^2+z^2+xy+yz+zx \leq 18$$
$$x^2+y^2+z^2+xy+yz+zx \leq 2\left(x^2+y^2+z^2\right) \leq 2 \cdot 9 =18.$$
Yes, it is ok :) 
thanks :)
 A: I try to apply Cauchy-Buniakowski and I obtaine the followin: 
$$\sum_{x,y,z}{\frac{1}{\sqrt{x^2+yz+3}}}\cdot \sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}\geq 9$$
So I have to prove that : $$\displaystyle\frac{9}{\sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}} \geq 1$$ if $x^2+y^2+z^2 \leq9$. 
$$\left(\sum_{x,y,z}{\sqrt{x^2+yz+3}}\right) \leq \sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)} $$ so we have to prove that: 
$$\frac{9}{\sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)}} \geq 1$$ hence:
$$3(x^2+y^2+z^2+xy+yz+zx+9) \leq 81$$ or
$$(x^2+y^2+z^2+xy+yz+zx+9) \leq 27$$ or
$$x^2+y^2+z^2+xy+yz+zx \leq 18$$
$$x^2+y^2+z^2+xy+yz+zx \leq 2\left(x^2+y^2+z^2\right) \leq 2 \cdot 9 =18.$$
A: I found an easier proof: Let $a=\sqrt{x^2+xy+3}$ and $b,c$ so on. Since $xy+yz+xz\leq x^2+y^2+z^2\leq 9$ we know that $a^2+b^2+c^2\leq 27$. Then $3(abc)^{\frac{2}{3}}\leq a^2+b^2+c^2\leq 27$, i.e. $(abc)^{\frac{1}{3}}\leq 3$. Then
\begin{align}
LHS&=\sum_{cyc}\frac{1}{a}\\
&=\frac{ab+bc+ac}{abc}\\
& \geq\frac{3(abc)^{\frac{2}{3}}}{abc}\\
&= 3(abc)^{-\frac{1}{3}}\\
&\geq 1.
\end{align}
