If $x,y,z>0$ with $x+y+z=\sqrt{3}$, prove that: $\sqrt{x^2+1-yz}+\sqrt{y^2+1-zx}+\sqrt{z^2+1-xy}\ge 3$ (Venezuela 1960) 
If $x,y,z>0$ with $x+y+z=\sqrt{3}$, prove that: $$\sqrt{x^2+1-yz}+\sqrt{y^2+1-zx}+\sqrt{z^2+1-xy}\ge 3$$

I tried solving it as follows:
The condition we want proved is: $\sqrt{3x^2+3-3yz}+\sqrt{3y^2+3-3zx}+\sqrt{3z^2+3-3xy}\ge 3(x+y+z)$
$\sqrt{3x^2+(x+y+z)^2-3yz}+\sqrt{3y^2+(x+y+z)^2-3xz}+\sqrt{3z^2+(x+y+z)^2-3xy}\ge 3(x+y+z)$
And this is where I got stuck. I couldn't continue it. Could you please explain to me how to solve the question?

The inequality ia actually very easy with just one hidden point kept in mind
Note that $$\sum \sqrt{x^2+1-yz}\ge \sum \sqrt{x^2+1-\frac{{(y+z)}^2}{4}}$$ Now use the fact $$x^2+1-\frac{{(y+z)}^2}{4}=x^2+\frac{{(x+y+z)}^2}{3}-\frac{{(y+z)}^2}{4}=\frac{1}{12} {(4x+y+z)}^2$$
