4 simultaneous equations in real numbers

Question

Solve the following system of equations in real numbers:

$\begin{cases}x^2+zx=y+z\\y^2+xy=z+x\\z^2+yz=x+y\\xyz=1\end{cases}$

Answer 1

Making $y = \nu x, z = \mu x$ we have

$$ x = \frac{\nu+\mu}{1+\mu} = \frac{\mu+1}{\nu(\nu+1)}=\frac{1+\nu}{\mu(\mu+1)} $$

Giving the two systems

$$ \frac{\nu+\mu}{1+\mu} = \frac{\mu+1}{\nu(\nu+1)}\\ \frac{\mu+1}{\nu(\nu+1)}=\frac{1+\nu}{\mu(\mu+1)} $$

and

$$ \frac{\nu+\mu}{1+\mu} = \frac{\mu+1}{\nu(\nu+1)}\\ \frac{\nu+\mu}{1+\mu}=\frac{1+\nu}{\mu(\mu+1)} $$

solving for $\nu, \mu$ we get as the solutions intersection

$$ \left[ \begin{array}{cc} \nu & \mu\\ 1 & 1 \\ \end{array} \right] $$

and the real solution is $\nu = \mu = 1$ with $x=y=z=1$

Attached a plot showing the three intervinent independent equalities.

Answer 2

Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.

Thus, $w^3=1$ and after summing of first three equations we obtain: $$\sum_{cyc}(x^2+xy)=2(x+y+z)$$ or $$9u^2-3v^2=6u,$$ which gives $$v^2=3u^2-2u.$$ Now, $$(x-y)^2(x-z)^2(y-z)^2\geq0$$ gives $$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0$$ or $$3u^2(3u^2-2u)^2-4(3u^2-2u)^3-4u^3+6u(3u^2-2u)-1\geq0$$ or $$(u-1)^2(81u^4-18u^3+15u^2+2u+1)\leq0$$ and since $$81u^4-18u^3+15u^2+2u+1=u^2(9u-1)^2+(u+1)^2+13u^2>0,$$ we obtain $u=1$, $v^2=1$ and since $x$, $y$ and $z$ they are roots of the equation $$(t-x)(t-y)(t-z)=0$$ or $$t^3-3ut^2+3v^2t-w^3=0$$ or $$t^3-3t^2+3t-1=0$$ or $$(t-1)^3=0,$$ we obtain $$x=y=z=1.$$ Easy to see that these values are valid and we are done!

There is also the following solution.

We'll prove that $$\sum_{cyc}(x^2+xy)\geq2(x+y+z)$$ for all reals $x$, $y$ and $z$ such that $xyz=1$.

Indeed, we need to prove that $$\left(\sum_{cyc}(x^2+xy)\right)^3\geq8xyz(x+y+z)^3,$$ which is a linear inequality of $w^3$, which says that it's enough to prove our inequality for an extreme value of $w^3$, which happens for an equality case of two variables.

Since the last inequality is even degree and homogeneous, it's enough to assume $y=z=1$, which gives $$(x-1)^2(x^4+8x^3+28x^2+44x+27)\geq0$$ and since $$x^4+8x^3+28x^2+44x+27=(x^2+4x+5)^2+2(x+1)^2>0,$$ our inequality is proven.

The equality occurs for $x=y=z$, which gives a solution of our system.

Answer 3

Let us give first the computer aided solution, sage. We associate the corresponding ideal, and ask for an elimination of two variables.

sage: R.<x,y,z> = PolynomialRing(QQ)
sage: J = R.ideal( [x*(x+z)-(y+z), y*(y+x)-(z+x), z*(z+y)-(x+y), x*y*z-1] )
sage: J.elimination_ideal( [x,y] )
Ideal (z^7 + z^5 + z^4 + 3*z^3 - 7*z^2 + 2*z - 1) 
    of Multivariate Polynomial Ring in x, y, z over Rational Field

sage: factor( z^7 + z^5 + z^4 + 3*z^3 - 7*z^2 + 2*z - 1 )
(z - 1) * (z^6 + z^5 + 2*z^4 + 3*z^3 + 6*z^2 - z + 1)

Now it turns out that the remained polynomial has exactly one real root, $z=1$, all other roots are complex. To see there is no real root, we can write for the quotient $$ (z^6 + z^5 + 2z^4 + 3z^3 + 6z^2 - z + 1) $$ as $$ \underbrace{(z^6+z^5+z^4)}_{\ge 0} + \underbrace{(z^4+3z^3+5z^2)}_{\ge 0} + \underbrace{(z^2-z+1)}_{> 0} >0\ . $$ From this we have solved the problem by symmetry. (We have so far $z=1$, same argument gives by cyclic permutation of variables also $y=1$ and $x=1$, and $x=y=z=1$ is a solution of the algebraic system.)

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Humanly, we have to work algebrically, but then it is hard to isolate an inequality with more than two variables, so the above elimination may be needed, or we search for some convexity argument for a function in three variables. Here is a way to eliminate as a human.

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We multiply the given equations $$ \left\{ \begin{aligned} x^2+zx &= y+z\\ y^2+xy &= z+x\\ z^2+yz &= x+y\\ xyz &=1 \end{aligned} \right. $$ with $1,1,x,-1$, add, get an equation of degree one in $x$, $$ x(z^2+z-1) = -y^2+y+2z-1\ . $$ From the equations and the above relation: $$ x = \frac{-y^2+y+2z-1}{z^2+z-1} =\frac{z-y^2}{y-1} =\frac 1{yz}\ . $$ We forget about $x$. Consider now the last equality as a polynomial equation of degree two in $z$, and use the other one to obtain a polynomial identity, then build the rest w.r.t. to the polynomial equation in $z$. The rest / reminder gives the elimination.

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Later edit. This algebraic solution i started is the "standard solution", it is not "beautiful", but it applies in all similar cases, we use dry elimination, and refuse to be "creative". From $$ \frac{-y^2+y+2z-1}{z^2+z-1} =\frac 1{yz} \text{ and } \frac{z-y^2}{y-1} =\frac 1{yz} $$ we obtain the two equations: $$ \begin{aligned} -y^3z + y^2z + 2yz^2 - yz - z^2 - z + 1 &=0\ ,\\ (*)\qquad yz^2 - y^3z - y + 1 &=0\ . \end{aligned} $$ (Now we build the resultant of the above polynomials of the L.H.S., seen as polynomials in $z$, or we work ad-hoc as follows.) From the second equation, we extract $\displaystyle z^2 = y^2 z + 1 +\frac 1y$, and insert in the first equation. All $z^2$-terms will be replaced. (This is the same as a polynomial division with rest.) We get: $$ y^4z - y^2z + 2y^2 - yz - 2y + 1=0\ . $$ This gives a formula for $z$ in terms of $y$: $$ z=-\frac{2y^2 - 2y + 1}{y^4-y^2-y}\ . $$ We insert this value in the equation $(*)$, this leads to... (computation done here with the computer)

sage: Z = -(2*y^2 - 2*y + 1) / y / (y^3 - y - 1)
sage: factor( y*Z^2 - y^3*Z - y + 1 )
y^-1 * (y - 1)^2 * (y^3 - y - 1)^-2 * (y^6 + y^5 + 2*y^4 + 3*y^3 + 6*y^2 - y + 1)

The only way to make the numerator vanish is via $y=1$. This gives then $z=1$, and finally $x=1$.

Answer 4

I think I have something. Vieta's equations with $(x-1)(y-1)(z-1)$

Expanding gives roughly the same as adding the 3 equations together. Swapping in the above and cancelling terms we find:

$$X(x-1)+y(y-1)+z(z-1)=(x-1)(y-1)(z-1)$$

$$(X-1/2)^2+(y-1/2)^2+(z-1/2)^2=(x-1)(y-1)(z-1)+3/4$$

We read off (1,1,1)

Solutions lie on a spheres centered at (1/2,1/2,1/2) that intersect a hyperboloid defined on the right. Let R be a given radius from the center of the sphere.

$R^2=(x-1)(y-1)(z-1)+3/4$ If $x=y=z$, then $x=(r^2-3/4)^{1/3}+1$

The hyperboloid has axis the line where x=y=z.

More possible solutions can be found using cylindrical symmetry.