103 Trigonometry Problems

5. Solutions to Advanced Problems 165
by the power mean inequality. Summing the above identities from 1 to n yields
n∑
1
n∑
cot θ i ≤
(n − 1) 3/2
i=1
∑
i=1 j̸=i
a i
a j
=
1 ∑
(n − 1) 3/2
1≤i,j≤n
i̸=j
a i
a j
(∗∗)
again, because each ratio a i
a j
(∗∗) gives
√
n∑
n − 1 tan θ i ≥
i=1
∑
1≤i,j≤n
i̸=j
appears once. Combining inequalities (∗) and
a j
a i
=
from which the desired result follows.
∑
1≤i,j≤n
i̸=j
a i
a j
≥ (n − 1) 3/2
n∑
cot θ i ,
i=1
39. [Weichao Wu] One of the two inequalities
(sin x) sin x (cos x) cos x
is always true for all real numbers x such that 0