# Solution to a system of symmetric equations

After applying the Lagrange multiplier method, I got the following system of equations, which is quite symmetric:

$(x+y)^2 + (x+z)^2 = \frac{2}{3} \lambda x$

$(y+x)^2 + (y+z)^2 = \frac{2}{3} \lambda y$

$(z+x)^2 + (z+y)^2 = \frac{2}{3} \lambda z$

It is claimed that this system of equations is satisfied only when $x=y=z$. Could anyone explain why?

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Obviously, $x=y=z$ leads to the solutions $x = y = z = 0$ and $x = y = z = \frac {\lambda}{12}$. Now, let's assume $x \ne y \ne z$. Subtract second equation from first, third from first and third from second, so you'll get equivalent system \begin{align} (x-y)(x+y+2z) &= \frac 23 \lambda (x-y)\\ (x-z)(x+2y+z) &= \frac 23 \lambda (x-z)\\ (y-z)(2x+y+z) &= \frac 23 \lambda (y-z) \end{align} or \begin{align} x+y+2z &= \frac 23 \lambda\\ x+2y+z &= \frac 23 \lambda\\ 2x+y+z &= \frac 23 \lambda \end{align} since we assumed that $x \ne y \ne z$.

Comparing first and second, one may conclude that $$x+y+2z = x+2y+z$$ or equivalently $$y = z$$ Considering other two pairs one may conclude $$x = y = z$$ which is in contradiction with initial assumptions. So the only solutions are those when $x = y = z$.

## Update

As it was pointed out, one also may check case when $x = y$ but $z \ne x,y$. In this case consider first and third equations \begin{align} 4x^2 + (x+z)^2 &= \frac 23 \lambda x \\ 2(z+x)^2 &= \frac 23 \lambda z \end{align} Subtract first from second $$(z-x)(z+3x) = \frac 23 \lambda (z-x)$$ Since we're assuming $z \ne x$, then $$z+3x = \frac 23 \lambda$$ or explicitly $$z = \frac 23 \lambda - 3x$$ Substitute it to the any of previous two, let's say second equation $$\left(\frac 23 \lambda - 2x\right)^2 = \frac 13 \lambda \left( \frac 23 \lambda -3x\right)$$ and if you solve it, you'll get complex values for $x$. So, contradiction.

In the same manner you can assume $z=y$, $x \ne y,z$ and $z=x$, $y \ne x,z$.

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This does not cover the case that only two out of three are equal (e.g. $x=y$, $z\neq x$, $z \neq y$). – WimC May 2 '13 at 7:34
@WimC you're right, it doesn't. I'll fix it in a minute. – Kaster May 2 '13 at 7:42

Interpreted as $3$-vectors, the left hand side and the right hand side (without the $\lambda$-factor) must be dependent. That means that all $2 \times 2$ minors of the matrix

$$\begin{pmatrix} (x+y)^2+(x+z)^2 & x\\ (y+x)^2+(y+z)^2 & y\\ (z+x)^2+(z+y)^2 & z \end{pmatrix}$$

must vanish (or in other words the cross product of the colums must be zero). Let's consider the first one:

$$0 = y \left( (x+y)^2 + (x+z)^2 \right) - x \left((y+x)^2 + (y+z)^2 \right) = (y-x)(x^2 + xy + y^2 + z^2).$$

So either $x=y$ or $x^2 + xy + y^2 + z^2 = 0$. The latter implies that $x=y=z=0$ since the quadratic form is positive definite over $\mathbb{R}$. In either case $x=y$. Other equalities follow similarly from the other minors.

One way to show that the quadratic form is positive definite is to write it as a positive combination of squares:

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

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What do you mean by "all 2×2 minors must vanish"? What matrix are you considering? – MichaelNgelo May 2 '13 at 18:19
@MichaelNgelo Added the matrix explicitly. – WimC May 2 '13 at 18:35