Solving the system $a^2-6=2\sqrt{2c+6}, \, b^2-6=2\sqrt{2a+6}, \, c^2-6=2\sqrt{2b+6}$ Question:

Solve the following system for $a,b,c\in \mathbb{R}$:
  $$\begin{cases}
b^2-6=2\sqrt{2a+6}\\
c^2-6=2\sqrt{2b+6}\\
a^2-6=2\sqrt{2c+6}
\end{cases}$$

I found the following:$$ (b^2-6)^2=4(2a+6)$$
$$(c^2-6)^2=4(2b+6)$$
$$(a^2-6)^2=4(2c+6)$$
Then maybe $a=b=c$ is one case.
Thank you.

Let $$ 3T:=a^2+b^2+c^2-18 = 2\sqrt{2a+6} +
   2\sqrt{2b+6} + 2\sqrt{2c+6} $$
If $$ a= {\rm max}\ \{ a,b,c\},\ b
 = {\rm min}\ \{ a,b,c\},\ a>b$$ then
 $$T>b^2-6=2\sqrt{2a+6} > T$$
Contradiction.
