Nice Inequality

It was derived from AoPS only.. I'm trying but could not get through it. So I thought it'll be better to share here too//

It looks nice, but when i do it, everything goes out of mind.

yes sambit, you can carry on from there.

\begin{align*} & tanx + tany + tanz = 1 \\ & \implies \Sigma \frac{sin^2x}{sin2x} = \frac{1}{2}...(*) \\ & \textup{now obviously }, \frac{sin^2x}{sin2x}\geq sin^2x (cyclic)\\ & \textup{so on adding all gives,}\\ & \Sigma \frac{sin^2x}{sin2x} \geq \Sigma sin^2x\\ & \textup{from .(*)}\\ & \boxed {\Sigma sin^2x \leq \frac{1}{2}} \\ & Q.E.D \\ & \\ & \end{align*}

x is in [0,90) →tanx≥0
(tanx-1)²â‰¥0
tan²x-2tanx+1≥0
2tanx≤tan²x+1
2tan²x≤tan³x+tanx
tan²x1+tan²x≤tanx2
→sin²x≤tanx2
sin²x+....≤tanx/2+.....=1/2

I think they had in mind:\because \ 1+\tan^2 x \ge 2 \tan x

\sin^2 x = \sum \frac{\tan^2 x}{1+\tan^2 x} \le \frac{1}{2} \sum \tan x = \frac{1}{2}

short and simple. thanks sir.