Question: I would appreciate if somebody could help me with the following problem

Q. Finding ~~maximum~~ minimum 
$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$

Answer: 
$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$

Consider the following : 

$a , \frac{1}{a}$ we know that $A.M. \geq G.M.$

$\therefore \frac{a+ \frac{1}{a}}{2} \geq \sqrt{a . \frac{1}{a}}$

$\Rightarrow \frac{a^2+1}{2a} \geq 1 $

$\Rightarrow a^2 + 1 \geq 2a $
$\Rightarrow (a-1)^2 \geq 0$

$\Rightarrow a \geq 1$ 

$\therefore $ the expression has only minimum value which is 1 and no maximum value. 

The expression $$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$

has minimum value of 1 + 1 +1 +1 +1 +1+1 = 6
A more formal proof that there is no maximum follows from taking the first derivatives and comparing to zero, showing that if $(x\_0, y\_0, z\_0)$ is a maximum / minimum, it satisfies: $$x\_0 = y\_0 = z\_0$$
This means that you can calculate the Hessian by only doing two calculations to determine $f\_{xx}$ and $f\_{xy}$, showing that at the maximum, say $(x\_0,x\_0,x\_0)$:
$$H=\frac{2}{x\_0^2}\left(\begin{matrix} 2 & 1 & 1\\
1 & 2 & 1 \\ 1& 1 & 2 \end{matrix}\right)$$
Since $H$ is clearly positive definite, the function has no maximum.