341are x,y,z>0?
In that case since tan x is convex in the interval, we have
\frac{\tan x + \tan y + \tan z}{3} \ge \tan \frac{x+y+z}{3} = \frac{1}{\sqrt 3}
so that
\tan x + \tan y + \tan z \ge \sqrt 3
since tan(x+y+z) is not defined we have tan x tany + tan y tan z + tan z tan x = 1 and hence from AM-GM we have
\tan x \tan y \tan z \le \frac{1}{(\sqrt 3)^3}
We will now prove that \sqrt 3 > 1+ \frac{1}{(\sqrt 3)^3}
This implies that (\sqrt 3)^4 - 1> (\sqrt 3)^3 or 8^2>3^3 which is true
Hence we have \tan x + \tan y + \tan z \ge \sqrt 3 > 1+ \frac{1}{(\sqrt 3)^3} > 1 + \tan x \tan y \tan z
