71It 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.
11I can only reach till
sin2xsin2x+sin2ysin2y+sin2zsin2z=12
Then I dont know how to proceed.
1yes 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*}
1x is in [0,90) →tanx≥0
(tanx-1)2≥0
tan2x-2tanx+1≥0
2tanx≤tan2x+1
2tan2x≤tan3x+tanx
tan2x1+tan2x≤tanx2
→sin2x≤tanx2
sin2x+....≤tanx/2+.....=1/2
341I 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}
