Exact solution to nonconvex optimization problem I would like to minimize $v+w+x+y+z$ subject to the following:
$$\frac{v+w}{x+y+z}=\frac{y}{z}=\frac{w}{x+y}$$
where $v,w,x,y,z\ge 1$
I tried entering this problem in WolframAlpha:
Minimize[{v + w + x + y + z, (v+w)/(x+y+z) == y/z &&  y/z== w/(x+y)&& v >= 1 && x >= 1 && y >= 1 && w >= 1 && z >= 1}, {v, w, x, y, z}]

It found a global min at $(v,w,x,y,z)=(1,\sqrt{2},1,1,\sqrt{2})$. What techniques might WolframAlpha be using to find this solution?
 A: *

*Simplifying the constraints. Set
$$
t=\frac{x+y+z}{v+w}=\frac{z}{y}=\frac{x+y}{w},
$$
then
$$
\left\{
\begin{array}{rcl}
x+y+z&=&t(v+w),\\
z&=&ty,\\
x+y&=&tw
\end{array}\right.
\quad\Rightarrow\quad y=v,
$$
and the problem becomes
$$
\min\, (1+t)(y+w)\quad\text{subject to }
\left\{
\begin{array}{rcl}
1&\le&ty,\\
1+y&\le&tw,\\
1&\le&y,\\
1&\le&w.
\end{array}\right.\quad\Leftrightarrow\quad
\left\{
\begin{array}{rcl}
1-ty&\le&0,\\
1+y-tw&\le&0,\\
1-y&\le&0,\\
1-w&\le&0.
\end{array}\right.\tag{*}
$$

*Applying the KKT necessary condition.
$$
\left\{
\begin{array}{rcl}
1+t-\mu_1t+\mu_2-\mu_3&=&0,\\
1+t-\mu_2t-\mu_4&=&0,\\
y+w-\mu_1y-\mu_2w&=&0,\\
\mu_1(1-ty)&=&0,\\
\mu_2(1+y-tw)&=&0,\\
\mu_3(1-y)&=&0,\\
\mu_4(1-w)&=&0,\\
\mu_k\ge 0\text{ and }(*)
\end{array}
\right.\quad\Leftrightarrow\quad
\left\{
\begin{array}{rcl}
1+(1-\mu_1)t+\mu_2-\mu_3&=&0,\quad(1)\\
1+(1-\mu_2)t-\mu_4&=&0,\quad(2)\\
(1-\mu_1)y+(1-\mu_2)w&=&0,\quad(3)\\
\mu_1(1-ty)&=&0,\quad(4)\\
\mu_2(1+y-tw)&=&0,\quad(5)\\
\mu_3(1-y)&=&0,\quad(6)\\
\mu_4(1-w)&=&0,\quad(7)\\
\mu_k\ge 0\text{ and } (*)&&\ \ \ \ \quad(8)
\end{array}
\right.
$$

*Solution. Equation (3) together with feasibility (*) leaves only 2 possibilites:
Case 1: $1-\mu_1\ge 0$, $1-\mu_2\le 0$ or
Case 2: $1-\mu_1\le 0$, $1-\mu_2\ge 0$.  


Consider those cases.
Case 1. We have $\mu_2>0$ and from (1) $\mu_3>0$, hence, from (5) and (6) it follows that $$y=1,\quad tw=1+y=2.$$
 - Assume $\mu_1=0$. Then from (3) we have $(\mu_2-1)w=1$ and from (2)
   $\mu_4=1-t/w$.
If $\mu_4=0$ then $t=w=\sqrt{2}$ and the objective value is $\color{red}{(1+\sqrt{2})^2}$.
If $\mu_4>0$ then from (7) $w=1$, thus $t=2$ and the objective value is $\color{red}{6}$.
- Assume $\mu_1>0$. Then from (4) we get $1=ty=t$, thus $w=2$ and the objective value is $\color{red}{6}$.
Case 2. We have $\mu_1>0$ and from (2) $\mu_4>0$, hence, from (4) and (7) it follows that
$$
w=1,\quad ty=1.
$$
 - Assume $\mu_2=0$. Then from (3) we get $(\mu_1-1)y=1$ and from (1) $\mu_3=1-t/y$.
If $\mu_3=0$ then $t=y=1$ and the objective value is $\color{red}{4}$.
If $\mu_3>0$ then from (6) $y=1$, hence, $t=1$ and the objective value is $\color{red}{4}$.
 - Assume $\mu_2>0$. Then from (5) $1+y=tw=t$. It gives $1=ty=(1+y)y$ $\Rightarrow$ $y=\frac{1}{1+y}\le \frac12$. It is not possible as it contradicts $y\ge 1$.  
Comparing the red objective values, it is clear that the smallest is for $y=1$, $t=w=\sqrt{2}$. Finally, $x=tw-y=1$, $z=ty=\sqrt{2}$ and $v=y=1$.
