It is well known that if $x + y + z = \pi$ then $$\tan x \times \tan y \times \tan z = \tan x+ \tan +\tan z.$$
I came across the following generalization of this equality:
$$\sqrt{1-k^2} {\rm sc}(x|k)\times {\rm sc}(y|k)\times{\rm sc}(z|k)\geq {\rm sc}(x|k)+ {\rm sc}(y|k)+{\rm sc}(z|k), $$ where ${\rm sc}(x|k)$ is the elliptic tangent function (by definition ${\rm sc}(x|k)=\frac{{\rm sn}(x|k)}{{\rm cn}(x|k)}$, where ${\rm sn}$ and ${\rm cn}$ are the Jacobi sine and cosine elliptic functions), and where $x+y+z=2K(k)$ ($K(k)$ is the generalization of $\pi/2$, it equals the complete elliptic integral of the first kind).
I already have an "analytic" proof of this inequality, but I would like to see a geometric proof. I am interested in any ideas!!