14 1. Preliminary Notions
Proposition 1.16. (AGM inequality) Given positive numbers x1,...,xr,
(1.18)
r
i=1
xi
1/r
≤
1
r
r
i=1
xi ,
where equality holds if and only if x1 = x2 = · · · = xr.
Proof. We give here a proof due to P´ olya. It is based on the inequality
ex
≥ 1 + x which is easily shown to be valid for all x ∈ R, with equality
if and only if x = 0. Given x1,...,xr ∈ R+, let M and P stand for their
arithmetic and geometric mean, respectively. If x1 = x2 = . . . = xr, then
M = P and we are done with the last claim. Therefore, it remains to prove
that P M if we assume that the xi’s are not all equal. Note that M 0.
In the inequality
ex
≥ 1 + x, replace x by xi/M − 1, so that
(1.19) exp
xi
M
− 1 ≥
xi
M
for i = 1, 2,...,r.
Now, since the xi’s are not all equal, it follows that xj M for at least one
j ∈ [1,r], implying that xj/M − 1 0. This means that at least one of the
inequalities in (1.19) is strict. Therefore, multiplying all the inequalities in
(1.19), we get
r
i=1
exp
xi
M
− 1
r
i=1
xi
M
,
exp
1
M
r
i=1
xi −
r
i=1
1
1
M
r
r
i=1
xi ,
er−r
P
r
M r
,
implying that M P , which is what we wanted to prove.
Proposition 1.17. (Cauchy-Schwarz inequality) Given two sets of real num-
bers a1,...,ak and b1,...,bk,
(1.20)
k
i=1
aibi
2
≤
k
i=1
ai
2
k
i=1
bi
2
.
Proof. This result can be proved by noticing that the quadratic polynomial
t2
k
i=1
ai
2
− 2t
k
i=1
aibi +
k
i=1
bi
2
=
k
i=1
(ait −
bi)2
