When $xyz=1$ why is $x+y+z\geq3$? ($x,y,z>0$) How can I prove the statement below without using the Inequality of arithmetic and geometric means?
$\forall x,y,z \in F  $ (F is an ordered field) $(x,y,z>0)$ $xyz=1 \implies x+y+z\geqslant3$
For the case where $x,y,z = 1$ it's easy to understand, but I fail to grasp how many more cases of $x,y,z$ ,I have to prove that $x+y+z\geqslant3$ work's for.
 A: Let $x\geq1$ and $y\leq1$.
Thus, $$(x-1)(y-1)\leq0$$ or
$$x+y\geq xy+1.$$
Thus, $$x+y+z\geq xy+1+z$$ and it's enough to prove
$$xy+z\geq2$$ or
$$xyz+1-xy-z\leq0$$ or
$$(xy-1)(z-1)\leq0,$$ which is obvious.
A: Cauchy/reverse/forward-backward induction works in any ordered field:
Start with $$2^2ab \leq (a+b)^2$$ and deduce: (forward induction) $$4^4xyzu \leq (x+y+z+u)^4$$ by letting $a = x+y$, $b = z+u$.
Then let $u = \frac{x+y+z}3$ to get: (backward induction) 
$$\frac{4^4}3xyz(x+y+z) \leq \left(\frac43\right)^4(x+y+z)^4$$
Finally $x,y,z>0$ so $x+y+z>0$ and we can divide by it:
$$3^3xyz \leq (x+y+z)^3$$
To finish, note that we can't take cube roots(*), so proceed by contradiction: if $x+y+z<3$, the above gives a contradiction.

(*) Under the additional assumption that $x,y,z$ are cubes, say $a^3,b^3,c^3$, we can simply use the identity
$$x+y+z-3 = \frac12(a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2\right)\geq0$$
A: Given $u=x+y+z$, use Lagrange multipliers:
$$L=x+y+z+k(1-xyz)$$
$$\begin{cases} L_x=1-kyz=0 \\ L_y=1-kxz=0 \\ L_z=1-kxy=0 \\ L_k=1-xyz=0 \end{cases} \Rightarrow x=y=z=k=1.$$
Bordered Hessian:
$$\begin{vmatrix} 0 & yz & xz & xy \\ 
yz & 0 & -kz & -ky \\ 
xz & -kz & 0 & -kx \\
xy & -ky & -kx & 0\end{vmatrix}=\begin{vmatrix} 0&1&1&1\\
1&0&-1&-1\\
1&-1&0&-1\\
1&-1&-1&0\end{vmatrix}.$$
$$\bar{H}_1=-1<0; \bar{H}_2=-2<0; \bar{H}_3=-3<0 \Rightarrow u(1,1,1)=3 \ (\text{min}).$$
