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 $x+y+z\geq \sqrt{3}$.

$(x+y+z)^2 \geq 3 \iff x^2+y^2+z^2 + 2(xy+yz+zx) \geq 3 \iff x^2+y^2+z^2+ \dfrac{2}{\sqrt{x^2+y^2+z^2}}\geq 3 \iff t^2 + \dfrac{2}{t} \geq 3$. By AM-GM inequality: $t^2+\dfrac{2}{t} =t^2+ \dfrac{1}{t}+\dfrac{1}{t}\geq 3$.
