103 Trigonometry Problems

5. Solutions to Advanced Problems 127
from which the desired result follows as a consequence of Introductory Problem
27(b).
Note: The above solution is similar to the proof of Chebyshev’s inequality.
We can also apply the rearrangement inequality to simplify our work.
Because a ≥ b ≥ c and cos A ≤ cos B ≤ cos C, wehave
a x cos A + b x cos B + c x cos C ≤ a x cos B + b x cos C + c x cos A
and
a x cos A + b x cos B + c x cos C ≤ a x cos C + b x cos A + c x cos B.
Hence
3(a x cos A + b x cos B + c x cos C)
≤ (a x + b x + c x )(cos A + cos B + cos C).
3. Let x,y,z be positive real numbers.
(a) Prove that
x
√
1 + x 2 +
y
√
1 + y 2 +
z
√ ≤ 3√ 3
1 + z 2 2
if x + y + z = xyz;
(b) Prove that
x
1 − x 2 + y
1 − y 2 + z
1 − z 2 ≥ 3√ 3
2
if 0