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 
$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$
 A: $$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$
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 
A: 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: 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.
A: 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.  
