I have the following constraint that I want to set in CVX:

A+B-C < sqrt(-A^2+(2B+2C)*A-(B-C)^2)

The lhs is affine hence convex.
I also know that the variables A, B, C are always positive.
I proved that under the positivity assumption, the rhs of the inequality above is concave, meaning that the constraint is convex.

How do I formulate the sqrt function on the rhs such that CVX will accept it as a concave function?

I implicitly assume that what’s inside the sqrt is always positive.
I’m claiming that the inequality constraint defines a convex region in R^3 wherever it’s well defined.
Under this assumption I’m looking for a way to make CVX accept the constraint.

If it makes the answer to my question easier one can further assume the following stronger conditions on the domain of definition:

\sqrt A + \sqrt B > \sqrt C

\sqrt A + \sqrt C > \sqrt B

\sqrt B + \sqrt C > \sqrt A

These inequalities ensure that the sqrt is always well defined.
A=1, B=0, C=0 leads to a negative sqrt. It is contradicted by 3).
These constraints define a convex region.
For example, since A,B,C are positive, 3) can be rewritten as: 2\sqrt{BC} > A-B-C, which is convex since \sqrt{BC} is concave.

then you have a rotated quadratic cone (rotated_lorentz) assuming A>0, B+C>0.

EDIT: To be completely equivalent with the original formulation (before squaring) we should also account for the case when LHS is strongly negative, more so than -RHS. I guess this should do:

In CVX, I believe this would be {[A+B+C; A; B-C],2*A,B+C} = rotated_lorentz(3)

Edit: See typo corrections by @Convexy below, one of which (A+B+C should be A+B-C) was also a “reado” on the original problem, necessitating the correction above by @Michal_Adamaszek