103 Trigonometry Problems

5. Solutions to Advanced Problems 149
We have
Likewise, we have
and
|EF| ≥|E 1 F 1 |=a − (|BF| cos B +|CE| cos C).
|DE| ≥c − (|AE| cos A +|BD| cos B)
|FD|≥b − (|CD| cos C +|AF | cos A).
Note that |DC| +|CE| =|EA| +|AF |=|FB|+|BD| = 1 3
(a + b + c).
Adding the last three inequalities gives
|DE|+|EF|+|FD|
≥ a + b + c − 1 (a + b + c)(cos A + cos B + cos C)
3
≥ 1 (a + b + c),
2
by Introductory Problem 27(b). Equality holds if and only if the length of
segment EF (FD and DE) is equal to the length of the projection of segment
EF on line BC (FD on CA and DE on AB), and A = B = C = 60 ◦ , that
is, if and only if D, E, and F are the midpoints of an equilateral triangle.
27. Let a and b be positive real numbers. Prove that
1
√
1 + a 2 + 1
√
1 + b 2 ≥ 2
√ 1 + ab
if either (1) 0