Minimal surface in a ball Assume a minimal surface $\Sigma$ has boundary on the unit sphere in the Euclidean space
and $r$ is the distance from $\Sigma$ to the center of the ball.
Is it true that 
$$\mathop{\rm area} \Sigma\ge \pi\cdot(1-r^2).$$
Comments:


*

*The problem is solved in all dimensions and codimension, see "Area bounds for minimal..." by Brendle and Hung in 2016. (Thanks Rbega for the ref.)

*If $r=0$, the statement follows directly from the monotonicity formula.

*If $\Sigma$ is topological disc the answer is YES, see answer of Oleg Eroshkin below.

*The general question is formulated as a conjecture in 1975 --- see comment of Ian Agol.

*There is an analog in all dimension and codimension for area minimizing surfaces, see Alexander, H.; Hoffman, D.; Osserman, R. Area estimates for submanifolds of Euclidean space. 1974. 

 A: This has just been solved (in full generality) by Brendle and Hung using the first variation formula together with a clever (if mysterious) choice of vector field.
A: One obvious observation (of which you are probably already aware) is that if the boundary of the surface is connected, it must have length at least $2\pi\sqrt{1-r^2}$, or else it is contained in a lune whose convex hull does not contain a point at distance $r$ from the center. In the very special case that your surface is a topological disk transverse to a foliation of the ball by concentric spheres, the coarea formula (obtained by integrating the lengths of the intersection of your surface with concentric spheres, and using this observation) gives an estimate for the area, but a quick calculation shows that it is not good enough to prove what you want.
A: This result (and several similar) proved in a nice paper Alexander, H.; Osserman, R.
"Area bounds for various classes of surfaces."
Amer. J. Math. 97 (1975), no. 3, 753--769. 
A: I think there is such a minimal surface with area less than pi(1-r^2). Look at the catenoid formed by rotation of y a(cosh a) about the x axis where a is small the the distance from the origin will be 2a or more with the minimum at x=0. Now it will intersect the circle at a value of x less than (ln (a^-1))^2 actually well before that but clearly e^((log a)^2) is a^(ln a) (note this is a small number to a large negative power which gives a large number which is bigger than 2/a which is all I need (note I am also overestimating the length by assuming the radius where the catenoid intersects the sphere is one) I estimate the area with two cones using the formula 2(pi)rs which will have a greater area than the minimal surface and get an upper bound of 2(ln a)^2 - a small factor due to the circle at the origin which I ignore since the number is small enough already not to mention I am already grossly overestimating the length. So I get an upper bound of the area of 2(ln a)^2 which is less than pi(1-4a^2).
