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 $3\cdot2 = 6$.
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.