Finding minimum $\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}$ I would appreciate if somebody could help me with the following problem
Q. Finding maximum minimum 
$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$

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.  
