Determining the Existence of Global Minimum/Maximum Determine whether the function defined as $$f(x,y,z)=x+y+z$$ has a maximum or a minimum value on the set $xy+yz=1$, $xz+yz=4$, $x>0$, $y>0$, $z>0$. 

It is clear to me that it does have a minimum value as we have that  $f>{0}$; and calculations reveal that; $$f(\sqrt{3},2-\sqrt{3},2)=4$$ is a minimum value. Now, I suspect $f$ does not have a maximum value but I'm unsure how to properly show this. 

 A: So, $$1-xy=yz=1-zx\implies xy=zx\implies y=z$$ as $x\ne0$
So, $xy+yz=1\implies xy+y^2=1$
and $x+y+z=x+2y=\frac{1-y^2}y+2y=\frac{1+y^2}y=\frac1y+y$ 
Applying AM–GM inequality, $\frac1y+y\ge 2\sqrt{\frac1y\cdot y}=2$ as $y>0$
EDIT: due to the rectification of the Question
Eliminate $y$ from the given relations to get $zx^2+z^2x-4x-3z=0$ 
Now use Lagrange Multiplier
A: 
The constraint surfaces  $ \ y· (x + z) = 1 \ $ and $ \ z·(x+y) = 4 \ $ are skewed hyperbolic "cylinders", so they extend without limit in some directions.  The graph above shows the first of these, for which part of the surface asymptotically approaches the $ \ xz=$ plane ( $ \ y = 0 \ $ ) within the first octant ; the image at right illustrates the hyperbolic cross-section of the surface [in green] and the other asymptotic plane [in yellow], $ \ z = -x \ . $  In a similar manner (a graph is not included here), the surface $ \ z·(x+y) = 4 \ $ has the asymptotic planes $ \ z = 0 \ $ (the $ \ xy-$ plane) and $ \ y = -x \ . $  So the second surface approaches the $ \ xy- $ plane within the first octant.

The two constraint surfaces together (the first in green, the second in blue) intersect on a skewed hyperbola, marked in pale yellow, seen in the graph at left.  The branch seen to the right is outside of the first octant, but the "left" branch is entirely within it; this is seen more evidently in the graph on the right.  This part of the space curve has small positive values for $ \ y \ , $ but extends indefinitely in the $ \ x-$ direction as $ \ z \ $ approaches zero, or vice versa.  So the function $ \ f(x,y,z) = x + y + z \ $ on this portion of the space curve has no global maximum.
The system of Lagrange equations
$$ 1 \ = \ \lambda · y \ + \ \mu · z \ \ , \ \ 1 \ = \  \lambda·(x+z)  \ + \ \mu · z \ \ , \ \ 1 \ = \ \lambda · y \ + \ \mu·(x+y) $$
leads to $$ \lambda · (x-y+z) \ = \ 0 \ \ , \ \ \mu · (x+y-z) \ = \ 0 \ \ , \ \ \lambda · (x-y+z) \ = \ \mu · (x+y-z)  \ \ , $$
for which the only consistent point solution in the first octant is $ \ ( \  \sqrt{3} \ , \ 2 - \sqrt{3} \ , \ 2 \ ) , $ as you found (and as is indicated in the above right graph).  So the function $ \ x + y + z \ $ has a global minimum of $ \ 4 \ $ and no global maximum.
