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