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 $3\cdot2 = 6$.
Note that if $F(x,y,z) = \frac{x+y}{z}+\frac{x+z}{y} + \frac{y+z}{x}$, then $F(kx,ky,kz)=F(x,y,z),\ k>0$. So we will use 
Lagrange multiplier method. Let $g(x,y,z)=x+y+z$. Constraint is $x+y+z=1$. 

$$\nabla F = (\frac{1}{z}+\frac{1}{y} - \frac{y+z}{x^2},\frac{1}{z}+\frac{1}{x} - \frac{x+z}{y^2},\frac{1}{y}+\frac{1}{x} - \frac{x+y}{z^2} ) =\lambda \nabla g$$

So $$ \frac{x^2(z+y) -(z+y)^2}{x^2yz}=\frac{z^2(x+y) -(x+y)^2}{xyz^2}= \frac{y^2(z+x) -(z+x)^2}{xy^2z} =\lambda $$

$$ \frac{x^2(1-x) -(1-x)^2}{x^2yz}=\frac{z^2(1-z) -(1-z)^2}{xyz^2}= \frac{y^2(1-y) -(1-y)^2}{xy^2z} =\lambda $$

Note that $\lambda\neq 0$ by computation.

Hence we have $$ (xz-xyz-1)(x-z)=(xy-xyz-1)(x-y)=(yz-xyz-1)(y-z)=0$$

$x=z\neq y$ implies that $2x^3-3x^2+x-1=0$. But it has only one solution larger than $1$. 

If $x,\ y,\ z$ are distinct, $xz=xy=yz$. Contradiction.

So $x=y=z$. $F(1/3,1/3,1/3)=6$ is minimum.