Find the maximum of the value $x+y+z$ if such condition $n(x+y+z)=xyz$ Let $n$ be give postive intger, and $x\ge y\ge z$ are postive integers,such
$$n(x+y+z)=xyz$$
Find the $(x+y+z)_{\max}$
I have see this problem   only answer is $(n+1)(n+2)$,iff $x=n(n+2),y=n+1,z=1$
and How do it? 
 A: SUMMARY: it really is true that
$$(x+y+z)_{\max} = n^2 + 3n + 2,$$
this occurring only when $z=1, y=n+1, x=n^2 + 2n $
This works. I will leave $n=1,2,3$ as exercises for the reader. We take $n \geq 4.$ We are given $x \geq y \geq z \geq 1,$ with
$$  xyz = nx + ny + nz, $$
$$ x y z^2 = nxz + nyz + n z^2,  $$
$$ z^2 xy - n zx - nzy = n z^2,  $$
$$ z^2 xy - nzx - nzy + n^2 = n^2 + n z^2,  $$
$$ (zx - n)(zy-n) = n(n+z^2).  $$
First, if $z \geq n,$ then $x,y,z \geq n.$ We find $(zx-n)(zy-n) \geq (z^2 -n)^2.$ Therefore
$$ (zx - n)(zy-n) - n(n+z^2) \geq z^4 - 3 n z^2  = z^2(z^2 - 3n).  $$
The factor $z^2 - 3n$  is positive for $ z \geq n \geq 4.$ ADDENDUM When we include $n=1,2,3,$ we still get the conclusion $z \leq n$ from the same calculation. If we assume $z \geq n+1,$ we get a contradiction because $ z^2 - 3n \geq n^2 + 2 n + 1 - 3n = n^2 - n + 1 \geq \frac{3}{4} > 0.$ Therefore, one may finish $n=1$ with $z=1,$ $n=2$ with $z=1,2,$ and $n=3$ with $z=1,2,3.$
We continue with $n \geq 4$ and $z < n.$ To get to the punchline, the largest possible value with such fixed $n,z$ is when $z=1,$
 as suggested by the OP and my computer run last night.
Sketch of proof for fixed $1 \leq z < n.$ We have $$ (zx - n)(zy-n) = n(n+z^2).  $$ If both factors on the left hand side are non-positive, that means 
$ z(x+y) < 2n,  $ or $x+y \leq 2n,$ whence $x+y+z \leq 3n.$ This is small, we can do better. When both factors are positive, in particular we have $zy > n.$ Let
$$  n \equiv \delta \pmod z, $$
with $$ 0 \leq \delta < z. $$
Then $$ n + (z - \delta) \equiv 0 \pmod z.  $$
AUDIENCE REQUEST: if I have positive real numbers $AB=C,$ with fixed $C$ and lower bound $A\geq B \geq \epsilon > 0,$ the largest value of $A+B$ occurs when $B = \epsilon.$ This is calculus or Lagrange multipliers. To maximize $x+y,$ we are going to maximize $(zx - n)+(zy-n).$ These two summands have a fixed product $n(n+z^2),$ so the biggest sum occurs when $zy-n$ is as small as possible, that is $y$ is as small as possible. 
To minimize $y$ (check with Lagrange multipliers) we can take
$$ zy = n + (z - \delta) \leq n+z. $$ As a result,
$$ 1 \leq zy - n \leq z. $$  With
$$ (zx - n)(zy-n) = n(n+z^2),  $$
$$ (zx - n) \leq n(n+z^2).  $$ 
We have
$$ z^2 \leq z^2, $$
$$ zy \leq n + z,$$
$$ zx \leq n^2 + (z^2+1)n,$$
$$ z(x+y+z) \leq n^2 + (z^2+2)n + (z^2 + z) ,  $$
 $$ x+y+z \leq \frac{n^2 + (z^2+2)n + (z^2 + z) }{z} = \frac{(n+1)z^2 + z + n^2 + 2n }{z} = (n+1)z + 1 + \frac{ n^2 + 2n }{z} $$
 $$ x+y+z \leq  (n+1)z + 1 + \frac{ n^2 + 2n }{z} $$
The second derivative (in $z$) of the right hand side is positive, the first derivative of the right hand side is negative for small $z$  such as $1.$ The next value of $z$ for which the bound is as large as its value at $z=1$ is
$$ z = n + 1 - \frac{1}{n+1} > n.  $$
This means that with $z < n,$ the best value is when $z=1.$ Then $y = n+1$ and $x = n^2 + 2n.$
A: It's a hint, not a proof.
Be $x$ a polynomial of $n$ of degree $a$.  
Be $y$ a polynomial of $n$ of degree $b$. 
Be $z$ a polynomial of $n$ of degree $c$.
Be $z\le y\le x$ with $c\le b\le a$.
Then it has to be $1+\max(a,b,c)=a+b+c$.
This is only possible for $(b,c)=(1,0)$ with $a\ge 1$.
One gets conditions for the coefficients of the polynomials 
($a+1$ coefficients for $x$, $b+1$ coefficients for $y$, $c+1$ coefficients for $z$)
and therefore the number of equations have obviously to be $(a+1)+(b+1)+(c+1)=(a+1)(b+1)(c+1)$ and with $(b,c)=(1,0)$ one gets 
$a+4=2(a+1)$ which means $a=2$ and therefore $a\in \{1,2\}$.
Theese considerations are not exact, but they show the right direction.
I hope it helps.  
(In order to obtain a solution replace $x,y,z$ by polynomials of $n$ (degrees $(a,b,c)=(2,1,0)$) and calculate their coefficients.)
