Let $x,y,z$ be positive real numbers with $|x-y|<2, |y-z|<2, |x-z|<2$. Prove that $\sqrt{xy+1}+\sqrt{yz+1}+\sqrt{xz+1}>x+y+z$. Let $x,y,z$ be positive real numbers with $|x-y|<2, |y-z|<2, |x-z|<2$. Prove that $$\sqrt{xy+1}+\sqrt{yz+1}+\sqrt{xz+1}>x+y+z.$$
My approach: I have spent some time and came up with the following idea: Since $|x-y|<2$ then if we multiply by $y>0$ we will get $(y-1)^2<xy+1<(y+1)^2$ then it follows that $\sqrt{xy+1}>|y-1|$. And the same can be done for ech term and we will get that $\sqrt{xy+1}+\sqrt{yz+1}+\sqrt{xz+1}>|x-1|+|y-1|+|z-1|$. But I cannot get the desired inequality.
Can anyone show the solution please?

$|x-y|<2$ can be squared to give $x^2  -2xy + y^2<4$.
Add  $4xy$ which gives $x^2  +2xy + y^2<4 + 4 xy$.
Take the square root: $x+y<2 \sqrt{1 +  xy}$.
Adding this up for all 3 inequalities gives the result. $\qquad \Box$
