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I would appreciate if somebody could help me with the following problem

Q. Finding maximum minimum $$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$

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closed as off-topic by user223391, G Cab, hardmath, rtybase, Brian Borchers Feb 18 '18 at 18:53

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    $\begingroup$ The minimum is a little more interesting. $\endgroup$ – André Nicolas Oct 13 '13 at 2:38
  • $\begingroup$ It is worth mentioning that some of the answers below answer the original question, i.e. the question about maximum; before the question was edited. $\endgroup$ – Martin Sleziak Oct 13 '13 at 8:41
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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 $3\cdot2 = 6$.

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  • $\begingroup$ Simple and quite fine. Up vote $0$ k. $\endgroup$ – Felix Marin Oct 13 '13 at 10:07
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$$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}(\text{where} ~~x,y,z>0)$$

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

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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.

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  • $\begingroup$ Why the downvote? I still think this is correct. $\endgroup$ – nbubis Oct 14 '13 at 6:20
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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.

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  • $\begingroup$ The expression is invariant under permutations of $x$, $y$, and $z$, so the minimum occurs when they are equal. If $x=y=z$ then $F(x,y,z) = 6$. $\endgroup$ – Umberto P. Oct 13 '13 at 4:29
  • $\begingroup$ Symmetry implies that if critical point set exists, then it is invariant. I cannot convince that there is no invariant critical set as like circle. $\endgroup$ – HK Lee Oct 13 '13 at 7:21

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