Solve $\sqrt{a+bx}+\sqrt{b+cx}+\sqrt{c+ax}=\sqrt{b-ax}+\sqrt{c-bx}+\sqrt{a-cx}$ Let $a,b$ and $c$ be real and positive parameters. Solve the equation
$$\sqrt{a+bx}+\sqrt{b+cx}+\sqrt{c+ax}=\sqrt{b-ax}+\sqrt{c-bx}+\sqrt{a-cx}$$
What could I do? Should take the square of both sides?

Oleg567's comment & hint as answer.
$x=0\,$ is a (trivial) solution.
The LHS is increasing in $x$; the RHS is decreasing in $x$. So, $x=0$ is unique solution.
