103 Trigonometry Problems

5. Solutions to Advanced Problems 181
Viewing the given equality as a quadratic equation in a and solving for a yields
Note that
√
−bc ± b 2 c 2 − 4 ( b 2 + c 2) + 16
a =
2
.
b 2 c 2 − 4(b 2 + c 2 ) + 16 ≤ b 2 c 2 − 8bc + 16 = (4 − bc) 2 .
For the given equality to hold, we must have b, c ≤ 2, so that 4 − bc ≥ 0.
Hence,
−bc +|4 − bc| −bc + 4 − bc
a ≤ = = 2 − bc,
2
2
or
2 − bc ≥ a. (‡)
Combining the inequalities (†) and (‡) gives
2 − bc = (2 − bc) · 1 ≥ a(b + c − bc) = ab + ac − abc,
or ab + ac + bc − abc ≤ 2, as desired.
45. [Gabriel Dospinescu and Dung Tran Nam] Let s, t, u, v be numbers in the
interval ( 0, π 2
)
with s + t + u + v = π. Prove that
√ √ √ √
2 sin s − 1 2 sin t − 1 2 sin u − 1 2 sin v − 1
+
+
+
≥ 0.
cos s cos t cos u cos v
Solution: Set a = tan s, b = tan t, c = tan u, and d = tan v. Then a, b, c, d
are positive real numbers. Because s + t + u + v = π, it follows that tan(s +
t) + tan(u + v) = 0; that is,
a + b
1 − ab + c + d
1 − cd = 0,
by the addition and subtraction formulas. Multiplying both sides of the last
equation by (1 − ab)(1 − cd) yields
(a + b)(1 − cd) + (c + d)(1 − ab) = 0,
or
a + b + c + d = abc + bcd + cda + dab.