<s>[INST] 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 $x+y+z\geq \sqrt{3}$.
 [/INST] Taking the contrapositive, we suppose $x+y+z<\sqrt 3$ and want to prove
$$xy+yz+zx<\frac 1{\sqrt{x^2+y^2+z^2}}\iff (xy+yz+zx)^2(x^2+y^2+z^2)<1$$
Scale $(x,y,z)$ by dividing all variables by $\sqrt3$. Then $x+y+z<3$ and we want to prove
$$(xy+yz+zx)^2(x^2+y^2+z^2)<27$$
And that's a simple AM-GM:
$$(xy+yz+zx)^2(x^2+y^2+z^2)\le\left(\frac{2(xy+yz+zx)+x^2+y^2+z^2}3\right)^3=\left(\frac{(x+y+z)^2}3\right)^3<27$$
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