368 Algebra
Remark.In statistics the numbersfiare integers that record the frequency of occurrence
of the sampled random variablexi,i= 1 , 2 ,...,n.Iff 1 +f 2 +···+fn=N, then
s^2 =f 1 x 12 +f 2 x 22 +···+fnxn^2 −(f^1 x^1 +f^2 x^2 +···+fnxn)2
N
N− 1is called the sample variance. We have just proved that the sample variance is nonnegative.
107.By the Cauchy–Schwarz inequality,
(k 1 +···+kn)(
1
k 1+···+
1
kn)
≥n^2.We must thus have 5n− 4 ≥n^2 ,son≤4. Without loss of generality, we may suppose
thatk 1 ≤ ··· ≤kn.
Ifn=1, we must havek 1 =1, which is a solution. Note that hereinafter we cannot
havek 1 =1.
Ifn=2, we have(k 1 ,k 2 )∈{( 2 , 4 ), ( 3 , 3 )}, neither of which satisfies the relation
from the statement.
Ifn =3, we havek 1 +k 2 +k 3 =11, so 2 ≤ k 1 ≤ 3. Hence(k 1 ,k 2 ,k 3 ) ∈
{( 2 , 2 , 7 ), ( 2 , 3 , 6 ), ( 2 , 4 , 5 ), ( 3 , 3 , 5 ), ( 3 , 4 , 4 )}, and only( 2 , 3 , 6 )works.
Ifn=4, we must have equality in the Cauchy–Schwarz inequality, and this can
happen only ifk 1 =k 2 =k 3 =k 4 =4.
Hence the solutions aren=1 andk 1 =1,n=3, and(k 1 ,k 2 ,k 3 )is a permutation of
( 2 , 3 , 6 ), andn=4 and(k 1 ,k 2 ,k 3 ,k 4 )=( 4 , 4 , 4 , 4 ).
(66th W.L. Putnam Mathematical Competition, 2005, proposed by T. Andreescu)
108.One can check that geometric progressions satisfy the identity. A slick proof of
the converse is to recognize that we have the equality case in the Cauchy–Schwarz
inequality. It holds only ifaa^01 =aa 21 = ··· =an− 1 /an, i.e., only ifa 0 ,a 1 ,...,anis a
geometric progression.
109.LetP(x)=c 0 xn+c 1 xn−^1 +···+cn. Then
P(a)P(b)=(c 0 an+c 1 an−^1 +···+cn)(c 0 bn+c 1 bn−^1 +···+cn)
≥(c 0 (√
ab)n+c 1 (√
ab)n−^1 +···+cn)^2 =(P (√
ab))^2 ,by the Cauchy–Schwarz inequality, and the conclusion follows.
110.First solution:Ifa 1 ,a 2 ,...,anare positive integers, the Cauchy–Schwarz inequality
implies
(a 1 +a 2 +···+an)(
1
a 1+
1
a 2+···+
1
an)
≥n^2.