MHB An intriguing system of equation

[anemone]

Hi MHB,

I've recently come across a system of equation problem that I could not solve it fully, I don't know how to prove that system, the resulted polynomial has only three real roots.

Problem:

Solve for all real solutions in terms of $k$ for the system below:

$x(x^2+9k^2)=2z(x^2+k^2)$

$y(y^2+9k^2)=2x(y^2+k^2)$

$z(z^2+9k^2)=2y(z^2+k^2)$

It's no hard to see that when $x=y=z$, then we have the solutions:

$(x,\,y,\,z)=(-\sqrt{7}a,\,-\sqrt{7}a,\,-\sqrt{7}a)\stackrel{\text{or}}{=}(0,\,0,\,0)\stackrel{\text{or}}{=}(\sqrt{7}a,\,\sqrt{7}a,\,\sqrt{7}a)$

But I don't know what other conclusions that I could draw when I considered the condition where $x\ne y \ne z$ that could lead me to solve for this problem.

Wolfram has confirmed that those are the only real solutions the the given system, and I tried, using the knowledge that I borrowed from inequality, geometry, function, etc and nope, nothing helped. And here I am, hoping to gain some useful advice from the members to solve this system successfully.

Any help would be much appreciated.

Thanks.

 

[Nono713]

This is just a hunch but the cyclic nature of the system along with the only solution being $x = y = z$ suggests that $x < y$ implies $y < x$ and vice versa, and same for $(y, z)$ and $(z, x)$ (or the other way around) so that the only solution can be $x = y = z$. I haven't tried, but if you haven't tried that it seems like it may be fruitful.

 

[anemone]

Thanks Bacterius for your reply and yes, I think I was able to complete the solution based on your suggestion and here is my work:

First, if we rewrite the first given equation as $\dfrac{2z}{x}=\dfrac{x^2+9k^2}{x^2+k^2}$, we see that $\dfrac{2z}{x}>0$ for all real $x$ and $k$.

This means both $x$ and $z$ are either both positive or both negative. By the same token, we also have $x$ and $y$ are both positive or negative, and so are $y$ and $z$.

Since $x^2,\,9k^2,\,k^2$ are all positive and if we apply AM-GM inequality on those, we see that

$x^2+9k^2\ge 2(\sqrt{9x^2k^2})=2(3xk)=6xk$ and $x^2+k^2\ge 2(\sqrt{x^2k^2})=2(xk)=2xk$

Now, if we consider the case where $x$ and $z$ are both positive, we see that

$\dfrac{2z}{x}=\dfrac{x^2+9k^2}{x^2+k^2}\ge \dfrac{6xk}{2xk}=3$

$2z\ge 3x$

$\therefore z>x$

Similarly, from $\dfrac{2x}{y}=\dfrac{y^2+9k^2}{y^2+k^2}$, we have $x>y$ and this suggests $z>y$ but from $\dfrac{2y}{z}=\dfrac{z^2+9k^2}{z^2+k^2}$, we have $y>z$. This leads to a contradiction.

The other case for which all $x,\,y,\,z$ are negative works the same way and so, there is only one condition in the given system of equation that we need to consider, and that is when $x=y=z$.