Algebra 369Fora 1 =x 0 −x 1 ,a 2 =x 1 −x 2 ,...,an=xn− 1 −xnthis gives
1
x 0 −x 1+
1
x 1 −x 2+···+
1
xn− 1 −xn≥
n^2
x 0 −x 1 +x 1 −x 2 +···+xn− 1 −xn=n^2
x 0 −xn.
The inequality from the statement now follows from
x 0 +xn+
n^2
x 0 −xn≥ 2 n,which is rather easy, because it is equivalent to
(
√
x 0 −xn−n
√
x 0 −xn) 2
≥ 0.
Equality in Cauchy–Schwarz holds if and only ifx 0 −x 1 ,x 1 −x 2 ,...,xn− 1 −xnare
proportional tox 0 −^1 x 1 ,x 1 −^1 x 2 ,...,xn− 11 −xn. This happens whenx 0 −x 1 =x 1 −x 2 = ··· =
xn− 1 −xn. Also,
√
x 0 −xn−n/√
x 0 −xn=0 only ifx 0 −xn=n. This means that
the inequality from the statement becomes an equality if and only ifx 0 ,x 1 ,...,xnis an
arithmetic sequence with common difference 1.
Second solution: As before, letai=xi−xi+ 1. The inequality can be written as
∑n−^1i= 1(
ai+1
ai)
≥ 2 n.This follows immediately fromx+x−^1 ≥2.
(St. Petersburg City Mathematical Olympiad, 1999, second solution by R. Stong)
111.Because
1
sec(a−b)
=cos(a−b)=sinasinb+cosacosb,it suffices to show that
(
sin^3 a
sinb
+
cos^3 a
cosb)
(sinasinb+cosacosb)≥ 1.This is true because by the Cauchy–Schwarz inequality,
(
sin^3 a
sinb
+
cos^3 a
cosb)
(sinasinb+cosacosb)≥(sin^2 a+cos^2 a)^2 = 1.