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 $x+y+z\geq \sqrt{3}$.
Taking the contrapositive, we suppose $x+y+z&lt;\sqrt 3$ and want to prove
$$xy+yz+zx&lt;\frac 1{\sqrt{x^2+y^2+z^2}}\iff (xy+yz+zx)^2(x^2+y^2+z^2)&lt;1$$
Scale $(x,y,z)$ by dividing all variables by $\sqrt3$. Then $x+y+z&lt;3$ and we want to prove
$$(xy+yz+zx)^2(x^2+y^2+z^2)&lt;27$$

And that's a simple AM-GM:
$$(xy+yz+zx)^2(x^2+y^2+z^2)\le\left(\frac{2(xy+yz+zx)+x^2+y^2+z^2}3\right)^3=\left(\frac{(x+y+z)^2}3\right)^3&lt;27$$
Sketch indicates the truth of the proposition. Blue boxes : $x+y+z &gt; \sqrt{3}$.
Blue with purple centers   $$xy + yz + zx &gt; \frac{1}{\sqrt{x^2+y^2+z^2}}$$
The surfaces indicate where equality holds.