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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!!

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  • $\begingroup$ If you can share your analytic proof and give a link to the analytic and geometric proofs of the tangent identity, it would be easier for others to try to find an analogous geometric proof. $\endgroup$ Oct 7, 2014 at 13:56

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For a proof of the standard triple tangent identity, see here. For an analytic proof of the generalized identity, one can replace $z$ by $2K(k)-x-y$ and use the addition formulas for Jacobi elliptic functions.

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