Prove $x+y+z\geq3$ given $xyz=1$ and $0<x,y,z\in\mathbb{F}$

Question

Prove $x+y+z\geq3$ given $xyz=1$ and $0<x,y,z\in\mathbb{F}$. Secondly, show that $x+y+z=3\iff x=y=z=1$.

($\mathbb{F}$ is the ordered field)

I've tried applying the following lemma with no success yet:

If $0<a<1$ and $1<b$ where $a,b\in\mathbb{F}$ then $1+ab<a+b$

Answer 1

Let $D=(x+y+z)^3-27xyz$ and $S=x+y+z$. Then $$ 2D=(S+6z)(x-y)^2+(S+6x)(y-z)^2+(S+6y)(z-x)^2\geq 0 $$ and so $(x+y+z)^3\geq 27xyz=27$ and therefore $x + y + z\geq 3$. We have $x + y +z=3$ iff $D=0$ iff $x=y=z$.

Answer 2

Let $f(x,y,z)=x+y+z$, $g(x,y,z)=xyz$

Solve the equations $\bigtriangledown f=\lambda \bigtriangledown g \tag {1}$

$g(x,y,z)=1 \tag{2}$

$x, y, z \ge 0$

$\implies$ $1=\lambda yz\tag{*}$ $1=\lambda xz\tag{**}$ $1=\lambda xy\tag{***}$ $g(x,y,z)=1 \tag{****}$ $\implies$ $x=y=z=\lambda$

Appliying in (2) we get the desired result.