103 Trigonometry Problems

170 103 Trigonometry Problems
y 2 sin 2 nC + z 2 sin 2 nB − 2yzsin nB sin nC
= (y sin nC − z sin nB) 2 ≥ 0,
which is evident.
From the above proof, we note that the equality case of the desired
inequality holds only if (y sin nC − z sin nB) 2 ≥ 0, that is, if y sin nC =
z sin nB, or
if
y
sin nB =
z
sin nC
. By symmetry, the equality case holds only
x
sin nA = y
sin nB = z
sin nC .
It is also easy to check that the above condition is sufficient for the
equality case to hold.
(c) The extended law of sines gives that
R
a = 2 sin A and its analogous
forms for
R b and R c . Dividing both sides of the desired inequalities by R2
and expanding the resulting right-hand side yields
4(yz sin 2 A + zx sin 2 B + xy sin 2 C)
≤ x 2 + y 2 + z 2 + 2(xy + yz + zx),
or, by the double-angle formulas,
x 2 + y 2 + z 2
[
]
≥ 2 yz(2 sin 2 A − 1) + zx(2 sin 2 B − 1) + xy(2 sin 2 C − 1
=−2(yz cos 2A + zx cos 2B + xy cos 2C),
which is part (b) by setting n = 2.
By the argument at the end of the proof of (b), we conclude that the
equality case of the desired inequality holds if and only if
x
sin 2A =
y
sin 2B =
z
sin 2C .
(d) Setting x = xa 2 , y = yb 2 , and z = zc 2 in part (c) gives
or
a 2 b 2 c 2 (xy + yz + zx) ≤ R 2 ( xa 2 + yb 2 + zc 2) 2
,
16R 2 [ABC](xy + yz + zx) ≤ R 2 ( xa 2 + yb 2 + zc 2) 2
,
by Introductory Problem 25(a). Dividing both sides of the last inequality
by R 2 and taking square roots yields the desired result.