Proving that if $xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}}$, then $x+y+z\geq \sqrt{3}$ If $x, y, z$ are positive real numbers such that 
$$xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}},$$
then prove that $x+y+z\geq \sqrt{3}$.


Sketch indicates the truth of the proposition. Blue boxes : $x+y+z > \sqrt{3}$.
Blue with purple centers   $$xy + yz + zx > \frac{1}{\sqrt{x^2+y^2+z^2}}$$
The surfaces indicate where equality holds. 
