1 inequality....

are 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