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?
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Try with am-gm inequallity – Martín Vacas Vignolo Aug 03 '16 at 03:20
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How do it? becasue $x,y,z$ are postive integers,not real numbers, – math110 Aug 03 '16 at 03:21
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that is exactly the condition of am-gm inequality – Kenny Lau Aug 03 '16 at 03:23
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@communnites Uh, $\mathbb N \subset \mathbb R$, so the AM-GM inequality holds for the natural numbers as well? – Mathemagician1234 Aug 03 '16 at 03:27
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2@KennyLau,can you post your AM-GM inequality to answer this problem,Thanks – math110 Aug 03 '16 at 03:27
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@Mathemagician1234,I think can't $=$,because $x+y+z\ge 3\sqrt[3]{xyz}$,iff$x=y=z$,but for this postive intgers can't $x=y=z$ – math110 Aug 03 '16 at 03:29
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@KennyLau Your deleted answer gives a minimum value of x+y+z, not a maximum! lol – Mathemagician1234 Aug 03 '16 at 04:02
3 Answers
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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
Will Jagy
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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 03 '16 at 21:46
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@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 03 '16 at 22:47
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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 $x = n^2 + 2n.$
Will Jagy
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1Your 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 03 '16 at 16:15
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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 04 '16 at 11:01
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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.)
user90369
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