Prove $(xyz)^{1/3}\geq \frac{3}{1/x+1/y+1/z}$

Question

For $x,y,z>0$ Prove $(xyz)^{1/3}\geq \frac{3}{1/x+1/y+1/z}$

Attempt:
setting $f(x,y,z)=(xyz)^{1/3}$ and $h(x,y,z)=\frac{3}{1/x+1/y+1/z}$ we can see that $f,h$ are homog. function with degree of 1.therefore we can assume that $1/x+1/y+1/z=1$ and prove $f(x,y,z)\geq 3$
Using lagrange multipliers with $f$ and $1/x+1/y+1/z=1$ we get:
$\nabla f=\lambda \nabla g$ where $g=1/x+1/y+1/z-1$
$(\frac{1}{3x}f,\frac{1}{3y}f,\frac{1}{3z}f)=-\lambda(1/x^2,1/y^2,1/z^2)\to f=-3\frac{\lambda}{x}=-3\frac{\lambda}{y}=-3\frac{\lambda}{z}\to x=y=z=3$ and I got stuck here trying to explain how this point is the global minimum and not local.

any help?(edit-Without using the AM-GM inequality )

Answer 1

Use the AM-GM inequality.

$$\frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{3}\geq (\frac{1}{x}\frac{1}{y}\frac{1}{z})^\frac{1}{3}$$

Now, just switch the sides and take the reciprocals.