103 Trigonometry Problems

194 103 Trigonometry Problems
side and results in the equality atbtct = abct 3 = 1. Hence we may assume
without loss of generality that abc = 1. Then there exist positive real numbers
x,y,z such that a = x/y, b = z/x, c = y/z. The rearrangement inequality
gives
x 3 + y 3 + z 3 ≥ x 2 z + y 2 x + z 2 y.
Thus
as desired.
a
b + b c + c a = x2
yz + y2
zx + z2
xy = x3 + y 3 + z 3
xyz
≥ x2 z + y 2 x + z 2 y
xyz
= a + b + c,
= x y + y z + z x
Now we prove our main result. Note that
( )( )( )
4 cos 2 A 4 cos 2 B 4 cos 2 C = (8 cos A cos B cos C) 2 ≤ 1
by Introductory Problem 28(a). Setting a = 4 cos 2 A, b = 4 cos 2 B, c =
4 cos 2 C in the Lemma yields
( ) cos A 2
+
cos B
( ) cos B 2
+
cos C
establishing inequality (†).
( ) cos C 2
= a cos A b + b c + c a ≥ a + b + c
= 4(cos 2 A + cos 2 B + cos 2 C),
51. For any real number x and any positive integer n, prove that
n∑ sin kx
∣ k ∣ ≤ 2√ π.
k=1
Solution: The solution is based on the following three Lemmas.
Lemma 1 Let n be a positive integer, and let a 1 ,a 2 ,...,a n and b 1 ,b 2 ,...,b n
be two sequences of real numbers. Then
n∑
k=1
∑n−1
a k b k = S n b n + S k (b k − b k+1 ),
k=1