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?

• Try with am-gm inequallity – Martín Vacas Vignolo Aug 3 '16 at 3:20
• How do it? becasue $x,y,z$ are postive integers,not real numbers, – communnites Aug 3 '16 at 3:21
• that is exactly the condition of am-gm inequality – Kenny Lau Aug 3 '16 at 3:23
• @communnites Uh, $\mathbb N \subset \mathbb R$, so the AM-GM inequality holds for the natural numbers as well? – Mathemagician1234 Aug 3 '16 at 3:27
• @KennyLau,can you post your AM-GM inequality to answer this problem,Thanks – communnites Aug 3 '16 at 3:27

The suggested solution in the question does seem to give the largest sum for a given ratio $n:$

1           6                 3       2       1

2           8                 4       2       2
2          10                 5       4       1
2          12                 8       3       1

3           9                 3       3       3
3          10                 5       3       2
3          14                 7       6       1
3          15                 9       5       1
3          16                12       2       2
3          20                15       4       1

4          12                 6       4       2
4          15                10       3       2
4          18                 9       8       1
4          21                14       6       1
4          30                24       5       1

5          12                 5       4       3
5          14                 7       5       2
5          16                10       4       2
5          22                11      10       1
5          24                15       8       1
5          28                20       7       1
5          30                25       3       2
5          42                35       6       1

6          14                 7       4       3
6          16                 8       6       2
6          18                12       3       3
6          24                18       4       2
6          26                13      12       1
6          30                20       9       1
6          36                27       8       1
6          56                48       7       1

7          15                 7       5       3
7          18                 9       7       2
7          27                21       3       3
7          30                15      14       1
7          33                21      11       1
7          45                35       9       1
7          48                42       4       2
7          72                63       8       1

8          15                 6       5       4
8          16                 8       4       4
8          20                10       8       2
8          21                12       7       2
8          21                14       4       3
8          24                16       6       2
8          34                17      16       1
8          35                20      14       1
8          35                28       5       2
8          39                26      12       1
8          44                32      11       1
8          54                48       3       3
8          55                44      10       1
8          90                80       9       1

9          16                 6       6       4
9          18                 9       6       3
9          20                12       5       3
9          22                11       9       2
9          28                21       4       3
9          32                24       6       2
9          38                19      18       1
9          40                24      15       1
9          42                27      14       1
9          52                39      12       1
9          66                54      11       1
9          70                63       5       2
9         110                99      10       1

10          18                 9       5       4
10          24                12      10       2
10          24                16       5       3
10          42                21      20       1
10          42                35       4       3
10          48                32      15       1
10          48                40       6       2
10          78                65      12       1
10         132               120      11       1
• Your data is suggestive and inspires a solution for the case $z=1$. Solving $$x=n(y+z)/(yz-n)=n(y+1)/(y-n)$$ implies that $y>n$. The inequality $x\ge y$ in turn then implies $y\le2n$ (as strongly suggested by that data). Therefore $y$ is in the range $[n+1,2n]$. The sum $x+y$ is a decreasing function of $y$ in this interval, so attains its maximum at $y=n+1$. It should be possible to similarly handle the cases where $z>1$, but I can't think straight right now. It's also past midnight here... – Jyrki Lahtonen Aug 3 '16 at 21:46
• @JyrkiLahtonen, Enjoy Rio – Will Jagy Aug 3 '16 at 22:10
• @JyrkiLahtonen if you wake up feeling all enthusiastic, I made a separate answer with proof; it was most convenient to consider only $n \geq 4$ I think it was... – Will Jagy Aug 3 '16 at 22:47

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 $x = n^2 + 2n.$

• Your equation $(zx-n)(zy-n) = n(n+z)$ needs an explication, because it's not the same condition as in the question. Perhaps you mean $(zx-n)(zy-n) = n(n+z^2)$ ? – user90369 Aug 3 '16 at 16:15
• @user90369 fixed. – Will Jagy Aug 3 '16 at 19:33
• To minimize $y$ is an useful argumentation, but it's not possible to see where it comes from. Maybe it would be good to explain here a bit more. :-) – user90369 Aug 4 '16 at 11:01
• @user90369 paragraph added. – Will Jagy Aug 4 '16 at 17:51

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.)