When you bend a straight plate into a curved one with radius $r$, then the maximum stress will be:
$$\sigma_{max} = \frac{E}{1-\nu^2}\cdot \frac{z_{max}}{r}$$
where:
- $E$ is Youngs modulus
- $\nu$ is Poisson ratio
- $z_{max}$ is maximum distance from neutral axis to surface, which in case of pure bending is half of the plate thickness
(If the section is more like a beam (similar width and height), there is $1$ instead of the $1-\nu^2$, but I think that is not your case.)
So if your section has a rectangular shape with height of 1 mm ($z_{max} =$ 0.5 mm) and width of 10 mm, just the height is important for maximum curvature in elastic state but the width of course affects rigidity. If this is from a common structural steel with:
- $E = $ 200 000 MPa
- $\nu =$ 0.3
- $R_e =$ 235 MPa (yield stres)
You can use this with the first equation:
$$R_e = \sigma_{max} = \frac{E}{1-\nu^2}\cdot \frac{z_{max}}{r}$$
So the minimum radius for elastic bending is:
$$r = \frac{E}{1-\nu^2}\cdot \frac{z_{max}}{R_e} = 468 \text{ mm}$$
Such radius would be barely noticeable at the length of 15 mm.
You could almost halve the radius using titanium, which has comparable yield strength but lower Youngs modulus.
So if you aim for 10 mm radius you will definitely need to reduce the thickness $h$:
$$h_{max} = 2r\cdot R_e\cdot \frac{1-\nu^2}{E}$$
which for the structural steel is about 0.02 mm, which is 10 times thinner than a razor blade.
If you could have the insert on the outside instead of inside, you can increase the required radius and also the maximum thickness in elastic state. This option could be also safer.
Alternative solution for helping the fingers straighten could be to have guided tension springs on the outside of fingers. Or spiral springs on finger sides like here.
