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: 
By letting $u = \frac xy$ , $v = \frac yz$, and $w = \frac zx$ , our expression becomes $(u + \frac1u) + (v + \frac1v) + (w + \frac1w)$ , whose minimum is thrice that of $f(t) = t + \frac1t$ , which is to be found among the roots of its first order derivative: $f'(t) = 1 - \frac1{t^2}$ , which vanishes for $t = \pm1$ . Since *t* is positive, the only viable solution thus becomes $t = 1$ , for which $f(t) = 1 + \frac11 = 2$ , which yields a minimum value of $3\cdot2 = 6$.
$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$

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