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
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.