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.
On
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 $$x+y+z-3 = \frac12(a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2\right)\geq0$$
On
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: $$\begin{vmatrix} 0 & yz & xz & xy \\ yz & 0 & -kz & -ky \\ xz & -kz & 0 & -kx \\ xy & -ky & -kx & 0\end{vmatrix}=\begin{vmatrix} 0&1&1&1\\ 1&0&-1&-1\\ 1&-1&0&-1\\ 1&-1&-1&0\end{vmatrix}.$$ $$\bar{H}_1=-1<0; \bar{H}_2=-2<0; \bar{H}_3=-3<0 \Rightarrow u(1,1,1)=3 \ (\text{min}).$$
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.