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I have to solve $ x+y+z=1$ and $xyz=1$ for a set of $(x, y, z)$. Are there any such real numbers?

Edit : What if $x+y+z=xyz=r$, $r$ being an arbitrary real number. Will it still be possible to find real $x$, $y$, $z$?

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  • $\begingroup$ @SkeletonBow Link to the article. $\endgroup$ – nbro Dec 21 '16 at 13:25
  • $\begingroup$ It's not an article... $\endgroup$ – Skeleton Bow Dec 21 '16 at 13:25
  • $\begingroup$ @SkeletonBow Link to whatever thing that states it. $\endgroup$ – nbro Dec 21 '16 at 13:26
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    $\begingroup$ For a quick example, $(x,y,z)=(3,\frac 13\left(\sqrt6 -3\right),\frac 13\left(-3-\sqrt 6\right))$ $\endgroup$ – lulu Dec 21 '16 at 13:35
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    $\begingroup$ $x+y+z=1$ is the equation of a plane through three points $( 1,0,0), (0,1,0), (0, 0,1)$. Consider surface $xyz=1$ and try to find whether it crosses the plane or not. $\endgroup$ – Widawensen Dec 21 '16 at 15:20

15 Answers 15

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There are infinite solutions. $z=1-x-y$ is a plane's equation and $z=\dfrac{1}{xy}$ is a complicated curve, but their intersection presents infinitely many real points.

Intersection

At first I would try some trivial values like $x=0$ or $x=1$ or $x=-1$ and check which of them works and which doesn't.

For instance, if $x=0$, then $xyz=0$ doesn't work.

If $x=1$, $x+y+z=1\rightarrow y+z=0\rightarrow z=-y$ means that $xyz=1\rightarrow y^{2}=-1$ is a complex number also doesn't work.

Then $x=-1$ is a good attempt because $y+z=2$ and $yz=-1$ looks promising. \begin{equation*} yz=-1 \end{equation*} \begin{equation*} z=-\frac{1}{y} \end{equation*} then, \begin{equation*} y+z=2 \end{equation*} \begin{equation*} y-\frac{1}{y}=2 \end{equation*} \begin{equation*} y^{2}-1=2y \end{equation*} \begin{equation*} y^{2}-2y-1=0 \end{equation*} \begin{equation*} \Delta=4+4=8 \end{equation*} \begin{equation*} y=\frac{2\pm2\sqrt{2}}{2}=1\pm\sqrt{2} \end{equation*} \begin{equation*} \therefore z=1\mp\sqrt{2} \end{equation*}

Thus, both $\left(-1,1-\sqrt{2},1+\sqrt{2}\right)$ and $\left(-1,1+\sqrt{2},1-\sqrt{2}\right)$ are valid answers.

I could not comment above for the lack of space. Here are some notes about their signals:

As noticed before none of the variables can be zero, because $xyz\neq0$.

Also the three variables cannot be all positive, because if $x>0$, $y>0$, $z>0$, then it would put them under the interval $0<x,y,z<1$, if one of them would be greater than $1$, for example, $x>1$ then the sum $x+y+z$ would be greater than $1$, since $x>1$, $y>0$, $ z>0$. The problem of that interval is $0<y<1$ would mean that $\dfrac{1}{y}>1$ and for the same reason $0<z<1$ would mean $\dfrac{1}{z}>1$. But $xyz=1$ would make $x=\dfrac{1}{y}\cdot\dfrac{1}{z}>1$ and that would be impossible, so the three variables cannot be all positive and they cannot be localized on that interval.

On the other hand, you cannot have just one negative variable or three negative variables because their product would be negative, so you must have exactly two negative and one positive variables.

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    $\begingroup$ Just a pointer on language: not “infinite points”, but “infinitely many points”. $\endgroup$ – Lubin Dec 21 '16 at 18:03
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    $\begingroup$ "$z=1/{xy}$ is a complicated curve" - looks more like a surface to me. Or rather a 2-manifold with no boundary. $\endgroup$ – John Dvorak Dec 21 '16 at 18:35
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    $\begingroup$ @R.. This answer is addressing the former case, r = 1, and saying that if x = 0, then xyz=0, so no solution with x=0 works. $\endgroup$ – isaacg Dec 22 '16 at 9:05
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    $\begingroup$ Whatever software made that graph, it's making the surface of $xyz=1$ look much, much more complicated than it actually is. $\endgroup$ – David K Dec 22 '16 at 17:26
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    $\begingroup$ @David, then let me present a nicer image (also Mathematica generated). $\endgroup$ – J. M. is a poor mathematician Dec 23 '16 at 14:27
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$$x+y+z=1 \quad \text{and} \quad xyz=1$$

\begin{align} xy(1-x-y) &= 1 \\ xy - x^2y - xy^2 = 1 \\ xy^2 + x^2y - xy + 1 &= 0 \\ xy^2 + (x^2 - x)y + 1 &= 0 \\ y^2 + (x-1)y + \dfrac 1x &= 0 \qquad\text{{$x\ne 0$ since $xyz=1$.}}\\ y &= \dfrac 12(1-x) \pm \dfrac 12\sqrt{(x-1)^2-\dfrac 4x} \end{align}

So we need to find non negative real numbers, $n$, for which we can solve $$(x-1)^2-\dfrac 4x = n^2$$

Clearly then, there exists an $n$ for all negative values of x.

When x is positive, Wolfram alpha tells us that the single real-valued solution to $(x-1)^2-\dfrac 4x = 0$ is $x=\xi$, where $\xi =\dfrac 13 \left(2 + \sqrt[3]{53 + 6 \sqrt{78}} + \sqrt[3]{53 - 6 \sqrt{78}}\right) \approx 2.314596212276752$

A value of $n$ will exists for all $x \ge \xi$.


Working it out, the solution set is

$$\{x,y,z\} = \left\{x,\; \dfrac 12(1-x) + \dfrac 12\sqrt{(x-1)^2-\dfrac 4x},\; \dfrac 12(1-x) - \dfrac 12\sqrt{(x-1)^2-\dfrac 4x} \;\right\}$$

for all $x \in (-\infty, 0) \cup [\xi, \infty)$ where $\xi =\dfrac 13 \left(2 + \sqrt[3]{53 + 6 \sqrt{78}} + \sqrt[3]{53 - 6 \sqrt{78}}\right) \approx 2.314596212276752$

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    $\begingroup$ Simple, elementary, and direct. $\endgroup$ – DanielWainfleet Dec 21 '16 at 16:54
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    $\begingroup$ I may be missing something obvious, but I'm not seeing why you ruled out $n=0$. Can you expand on that? $\endgroup$ – hvd Dec 22 '16 at 10:51
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    $\begingroup$ A slightly different beginning would be to note that $zy = \frac 1x$ and $y+z = 1-x$. So $(w-y)(w-z) = w^2 -(1-x)w+\frac 1x$. Then $y,z = \dfrac{(1-x) \pm \sqrt{(1-x)^2 - \frac 4x}}{2}$ $\endgroup$ – steven gregory Dec 22 '16 at 12:19
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    $\begingroup$ @hvd - Yes, you're right. $n=0$ implies that $x=\xi$ and $y=z=\dfrac{1-\xi}{2} < 0$ $\endgroup$ – steven gregory Dec 22 '16 at 12:25
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Those numbers are the intersection of the surface $x y z=1$ and the plane $x+y+z=1$. A quick plot of these surfaces shows that these surfaces do intersect and that the solution consist of a curve (so infinitely many solutions) that looks something like a three-way hyperbola.

plot of surfaces

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  • $\begingroup$ What software is this? $\endgroup$ – ShreevatsaR Dec 21 '16 at 17:19
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    $\begingroup$ Wolfram mathematica (11.0 student edition). I used the function ContourPlot3d[] $\endgroup$ – user3502079 Dec 21 '16 at 17:25
  • $\begingroup$ I guess the asymptotes of the solution set are the permutations of $[t, -t, 0]$? $\endgroup$ – John Dvorak Dec 21 '16 at 18:39
  • $\begingroup$ If you use a mac the, built in graphing software makes nice images. Paremetizing the intersection of 2 surfaces could also be useful to OP $\endgroup$ – Prince M Dec 24 '16 at 3:05
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Suppose $x=-1/2$ then we have $y + z = 3/2$ and $yz=-2$. We have that $y,z$ are solutions to the quadratic $$a^2 - 3/2 a - 2.$$

Since the discriminant of this polynomial is positive there are $2$ real solutions (for $y$ and $z$). Hence there exists the required $x,y,z$.

Edit: Note that given any $x <0$ there is $y,z$ such that the $x,y,z$ satisfies your equations. The same proof that I used above works:

Given $x <0$, if $y+z=1-x > 0$ and $yz=\frac{1}{x} < 0$, then $y,z$ are solutions to the quadratic

$$a^2 -(1-x)a + \frac{1}{x}.$$

This quadratic has discriminant $(1-x)^2 -\frac{4}{x} >0.$ Hence there are solutions $x,y,z$ as required.

Second Edit: In the case we have $x + y + z = R$ and $xyz=S$ for $R,S \in \mathbb{R}$, $S\ \neq0$ (i.e a generalisation of your edit), we still have solutions. Using the same method as before, consider $y+z=R-x$ and $yz=S/x$. Then, $y,z$ are the roots to the polynomial

$$a^2 -(R-x)a + \frac{S}{x}.$$

As long as we fix $x$ so that $\frac{S}{x} <0$, then the discriminant of this quadratic is positive and we can guarantee a solution set $x,y,z$.

In the case $S=0$, one of $x,y,z$ must be $0$ - solutions are easy to see from there.

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Consider cubic

$$ (x-a)(x-b)(x-c) = x^3 - x^2( a+b+c) +x(a b+ b c+ca) -abc $$

Let $ d = (a b + b c+ca) $

$$ x^3 - x^2 + d\,x - 1 =0 $$

The roots satisfy your condition. But constant $d$ must be chosen so that all roots are real by considering its discriminant in

Cubic Equation Wiki

resulting simply as condition: $ d < -1.2275 $

EDIT1:

(Earlier error in discriminant $/ d $ )

$$ 4 d^3 - d^2 +18 d -31 <0 ; \quad d < -1.2275 $$

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    $\begingroup$ You might want to expand a bit about how the sum-of-the-roots and the product-of-the-roots are related to a polynomial's coefficients and how you used those properties to derive this polynomial. $\endgroup$ – Fixed Point Dec 21 '16 at 20:12
  • $\begingroup$ Also the case a+b+c = r = abc mentioned in the question could be investigated. $\endgroup$ – Gribouillis Dec 22 '16 at 18:25
  • $\begingroup$ @Gribouillis In my answer below it is investigated in fact even more general case.$a+b+c=r_1, abc=r_2$ $\endgroup$ – Widawensen Dec 22 '16 at 18:33
  • $\begingroup$ This is wrong. The derivative of a cubic having 2 real roots does not imply the cubic has 3 real roots, because the local minimum and maximum might be on the same side of the x axis. That's exactly what happens when $d>3$, for example for $d=4$ i.imgur.com/FnokKpg.png. And $4-\frac{12}{d}>0$ implies $d>3$ OR $d<0$. The $d>3$ generates no solutions and $d<0$ generates all the real solutions with an additional condition because the derivative of a cubic having two real roots is necessary but not sufficient for the cubic to have 3 real roots. $\endgroup$ – Sophie Dec 23 '16 at 11:54
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Suppose there actually is such a triple, $(x,y,z)$, of real numbers. Let $u$ be a variable and let's force $x$, $y$, and $z$ to be the roots of a polynomial. We would consider \begin{align*} 0 &= (u-x)(u-y)(u-z) \\ &= u^3 - (x+y+z)u^2 + (xy + xz + yz) u - (x y z) \\ &= u^3 - u^2 + (xy + xz + yz) u - 1 \\ &= u^3 - u^2 + A u - 1 \text{,} \end{align*} where we have used the abbreviation $A = xy + xz + yz$. If we let $A$ range over the complex numbers, we get three (generically complex) roots that satisfy your equation. Clearly, if $x$, $y$, and $z$ are real, $A$ is real, so we should restrict $A$ to the reals, but this isn't quite strong enough to make $x$, $y$, and $z$ real. (In fact, if $A$ is greater than about $-2.6$, two of the roots are complex. The exact bound for $A$ is the one real root of $4A^3 - A^2 - 18 A + 31$, the discriminant of the polynomial in $u$.)

The above says we could take $A = -3$ and then $x$, $y$, and $z$ are \begin{align*} -1, 1-\sqrt{2}, \text{ and } 1+\sqrt{2} \text{.} \end{align*} The sum of the two conjugates is $2$ and adding the first, the sum of all three is $1$. The product of the conjugates is $1 - 2 = -1$, since this is an example of a difference of two squares, and so the product of all three is $1$.

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Using Grapher, a few more plots (for $r=1$) to complement the other answers:

enter image description here

enter image description here

enter image description here

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  • $\begingroup$ This is a good drawing also for my answer.. $\endgroup$ – Widawensen Dec 22 '16 at 19:06
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$x+y+z=1$ defines a plane in $\mathbb{R}^3$.

That plane contains the point $P = (1,0,0)$ where the product equals $0$.

That plane also contains the point $Q = (3,-1,-1)$ where the product equals $3$.

The line segment $$\overline{PQ} = \{(1-t)P+tQ \,\bigm|\, 0 \le t \le 1\} = \{(1 + 2t,-t,-t) \, \bigm| \, 0 \le t \le 1\} $$ lies entirely in that plane. So by the intermediate value theorem, that line segment contains a point where the product equals $1$.

We can get more explicit, finding coordinates of the desired point $(1+2t,-t,-t)$ by setting $$(1+2t)(-t)(-t)=1 $$ and solving the resulting cubic equation $$2t^3 + t^2 - 1 = 0 $$ for a root in the interval $0 \le t \le 1$ (which exists, of course, by the intermediate value theorem). I used http://www.tiger-algebra.com to get $t \approx 0.657298088$, and so the point is, approximately, $$(2.314596176,-0.657298088,-0.657298088) $$

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Consider the polynomial $$P(t)=(t-x)(t-y)(t-z)=t^3-t^2+pt-1$$ Where $p=xy+xz+yz$. The discriminant of the general cubic ($ax^3+bx^2+cx+d$) is $\Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$, and $\Delta\geq 0$ if and only if all 3 roots are real. The discriminant of $P(t)$ is $\Delta=-4p^3+p^2+18p-31$. Solving the inequality $\Delta\geq 0$ you get $$p\leq\frac{1}{12}\left(1-\frac{217}{\sqrt[3]{6371-624\sqrt{78}}}-\sqrt[3]{6371-624\sqrt{78}}\right)\approx -2.6107$$ So whenever that inequality is satisfied the roots of $t^3-t^2+pt-1$ are real and their sum and product is 1.

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    $\begingroup$ One of you and @narasimham is wrong: you're saying $p \leq -2.6107$ and he's saying the equivalent of $p < 1/3$. $\endgroup$ – Mark Hurd Dec 22 '16 at 14:30
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    $\begingroup$ Thanks for pointing it out @MarkHurd. Some of the confusion is probably because Narasimham's $d=\frac{1}{p}$ but I'm not sure what's going on with the different sign. This graph suggests the $-2.6$ figure is correct i.imgur.com/MMoLlCm.png. $\endgroup$ – Sophie Dec 22 '16 at 14:44
  • $\begingroup$ Yes, on seeing other answers mention $-2.6$ I've upvoted you and downvoted @narasimham. $\endgroup$ – Mark Hurd Dec 22 '16 at 15:00
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If any are zero the product is zero. If all are positive all are less than 1 and the product is less than one so at least one is negative. If three or one are negative then the product is negative so two are negative and one is positive.

Wolog $x \le y < 0; z = 1-(x + y)$ or lets relable $y = -a; a > 0; x = -a - e; e \ge 0; z = a+ e + 1$

So we must solve $a(a+e)(1+a+e) = 1$ so $a(e^2 + (1+2a)e + a(1+a) - \frac 1a) = 0$ so $e = \frac{-1-2a \pm \sqrt{4a^2+4a + 1 - 4(a(1+a) - \frac 1a})}{2}= \frac{-1-2a + \sqrt{ 1 + \frac 4a}}{2}$ with the condition $a > 0$ and $1+ \frac 4a \ge (1 + 2a)^2$ or $a^3 + a^2 \le 1$ of which there are no doubt an infinite number of solutions.

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  • $\begingroup$ Isn't that z=2a+e+1? $\endgroup$ – steven gregory Dec 25 '16 at 3:09
  • $\begingroup$ Probably. Maybe. ....yeah. $\endgroup$ – fleablood Dec 26 '16 at 7:04
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General case: $r$ being an arbitrary real number.

The point $(r,0,0)$ belongs to the plane $x+y+z=r$.
The point $(x,y,z)=((r-a), \dfrac{a}{2} , \dfrac{a}{2} )$ also belongs to this plane because $(r-a)+ \dfrac{a}{2}+ \dfrac{a}{2}=r$.

  • Consider function $b=f(a)= (r-a) \dfrac{a}{2} \dfrac{a}{2}-r $.

    This is a continuous cubic function with the parameter $r$ so it has values from $+\infty$ to $-\infty$ and it is assured that this function crosses line $b=0$ so we have at least one real solution $a_1$ for the equation $(r-a) \dfrac{a}{2} \dfrac{a}{2} =xyz=r$

  • Therefore real solution for the set of equations in the question always exists .

See picture below for some cases of $r$

enter image description here

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  • $\begingroup$ In fact the system of equations $x+y+z=r_1$ and $xyz=r_2$ has always real solution even if $r_1\neq{r_2}$. There is no way for the plane described by the first equation to miss surfaces described by the second one. $\endgroup$ – Widawensen Dec 22 '16 at 18:02
  • $\begingroup$ The choice of a function $b=f(a)$ is equivalent to the choice of a line from the plane $x+y+z=r$. Other constructed function give other solution for the equation $xyz=r$, so this method after transformation of the equation $xyz=r$ into a cubic one give also a way of finding solutions for this set of equations with the known methods of solving cubic equations (see en.wikipedia.org/wiki/Cubic_function). The crucial here is that the coefficient standing by $a^3$ is never equal $0$ so the equation is cubic and it has always at least one real solution (sometimes there are three solutions) $\endgroup$ – Widawensen Dec 23 '16 at 8:50
  • $\begingroup$ Please be more careful before approving bogus edits. $\endgroup$ – Bill Dubuque Aug 4 '17 at 13:16
  • $\begingroup$ @BillDubuque Ok..., some errors always may happen.. $\endgroup$ – Widawensen Aug 4 '17 at 13:35
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As there are two equations in three unknowns, you a free to set one unknow.

Now solve

$$\begin{cases}x+y=1-z,\\xy=\dfrac1z,\end{cases}$$

which gives

$$x,y=\frac12\left(1-z\pm\sqrt{(1-z)^2-\frac4z}\right).$$

Solutions exist when the discriminant is non-negative,

$$(1-z)^2-\frac4z\ge0.$$

This is certainly true for values of $z$ above $3$ (more precisely the real root of $z^3-2z^2+z-4=0$), or negative.

For example $-\dfrac32+\sqrt2,-\dfrac32-\sqrt2,4$.

There is no solution with $x,y,z>0$.

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Given $x$ and the requirements $x + y + z = xyz = 1$, we can derive possible values for $y$ which in turn fully determines $z$.

Substituting $z = 1 - x - y$ in $xyz = 1$:

$$xy (1 - x - y) = 1$$ $$xy - x^2y - xy^2 = 1$$ $$(-x)y^2 + (x - x^2)y - 1 = 0$$

The quadratic formula now tells us that y is given by:

$$y = {-(x - x^2) \pm \sqrt{(x - x^2)^2 - 4x} \over -2x}$$ $$y = \frac{1-x}{2} \pm {\sqrt{x^4 - 2x^3 + x^2 - 4x} \over 2x}$$

Before tackling the domain, consider this:

$$y = \frac{1-x}{2} \pm R(x)$$ $$z = 1 - x - (\frac{1-x}{2} \pm R(x))$$ $$z = \frac{1 - x}{2} - (\pm R(x))$$

Thus the sign of the square root is irrelevant, as it merely swaps $y$ and $z$ around. Now all that remains is finding the domain for $x$ such that the solution is real. This means $x^4 - 2x^3 + x^2 - 4x > 0$. You can take the time to solve this by hand (it's actually a cubic), or plug it straight into your favourite CAS and find out it's:

$$x < 0 \vee x > \frac{2+\sqrt[3]{53-6\sqrt{78}}+\sqrt[3]{53+6\sqrt{78}}}{3}$$ $$X < 0 \vee x > 2.3146...$$

So the final parametric solution is:

$$R(x) = {\sqrt{x^4 - 2x^3 + x^2 - 4x} \over 2x}$$ $$\{x, y, z\} = \left\{x, \frac{1 - x}{2} - R(x), \frac{1 - x}{2} + R(x)\right\}$$

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Another approach: as already noted in other answers, in view of Viète's formulas, you are asking whether there exists $a\in\mathbb{R}$ such that $t^3-t^2+at-1=0$ has three real solutions.

Equivalently, you are asking whether there is an $a$ such that the graphs of $y=t^3-t^2-1$ and $y=at$ intersect in three points. Equivalently, you are asking whether there is a line through the origin that intersects the graph of the cubic $y=t^3-t^2-1$ in at least two points.

If you plot this curve it's not complicated to find one graphically, for instance use plot t^3-t^2-1 and 10t in [-4,4] in Wolfram Alpha:

enter image description here

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A negative fact after many positive solutions: you'll never find a solution with 3 positive numbers! In fact in that case AM-GM holds $$\frac{x+y+z}{3}\geq \sqrt[3]{xyz} $$ which would give $$\frac{1}{3} \geq 1$$

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