Inequalities

we have
tan ∂_i = sin∂₁cos ∂₁ = √1-cos²âˆ‚₁cos∂₁ = √cos²âˆ‚₁+ cos²âˆ‚₂ +.....+cos²âˆ‚_i-1 + cos²âˆ‚_i+1+.....+ cos²âˆ‚_n
cos∂₁

from m-th power inequality √x²+y² √2≥x+y2

so tan∂_i ≥ cos∂₁+ .....+ cos∂_i-1+ cos∂_i+1+...+ cos∂_ncos∂₁√n-1

summing the inequality for i from 1 to n we get

tan∂₁+...+ tan∂_n ≥ 1√n-1\sum_{}^{}{1\leq i,j\leq n,i\neq j}a_ja_i

because each ratios a_ja_i appear only once

on the other hand

cot∂_i = cos∂_isin∂_i= cos∂_i√cos²âˆ‚₁+ cos²âˆ‚₂ +.....+cos²âˆ‚_i-1 + cos²âˆ‚_i+1+.....+ cos²âˆ‚_n

again we have from m-th power inequality

√x²+y² √2≥21/x + 1/y

so cot∂_i ≤ cos∂_i (1cos∂₁ + ....+1cos∂_i-1+ 1cos∂_i+1+...+ 1cos∂_n)(n-1)√n-1

again after summing up we get

cot∂₁+ ...+ cot∂_n ≤\sum_{}^{}{1\leq i,j\leq n,i\neq j}a_ja_i
(n-1)³²

combining above two inequality

we get

√(n-1) (tan∂₁ +....+ tan∂₂) ≥ \sum_{}^{}{1\leq i,j\leq n,i\neq j}a_ja_i ≥ (n-1)³²(cot∂₁ +....+ cot∂_n)

so (tan∂₁ +....+ tan∂_n) ≥(n-1)(cot∂₁+......+ cot∂_n)

Where can I get more details/applications of this Inequality.

@Subho BTW, Did u appear for this year's RMO?

i did, didn qualify ,will give a best attempt this time

"where can i get......"

please specify..

where u get such questions , i mean any book?

we know AM of mth power ≥ m-th power of AM

so x² + y²/2 ≥ [( x+y)/2 ]²

taking square root we get √x²+y²√2≥x+y2

from AM≥ HM

x+y2≥21/x+1/y

so √x²+y²√2 ≥ 21/x+1/y

GOOD!!

for 2nd

put
3z-1= -+{root i}[z-2]

that shud help