Advanced High-School Mathematics

158 CHAPTER 3 Inequalities

y = ln x

x

y

P

Q

Proof. The proof involves a ge-
ometrical fact about the graph of
the function y = lnx, namely
that for any two points P, Q
on the graph, the straight line
segment on the graph is com-
pletely below the graph.^5 Thus,
let P = P(ap,ln(ap)) and Q =
Q(bq,ln(bq)) be two points on the
graph ofy = lnx. For any value
of the parametert, 0 ≤t≤1, the
pointX=X(tbq+ (1−t)aq,tln(bq) + (1−t) ln(ap)) is a point on the
line seqment PQ. Since the graph of y = lnx lies entirely above the
line segmentPQ, we conclude that

ln(tbq+ (1−t)ap)≥tln(bq) + (1−t) ln(ap) =tqlnb+ (1−t)plna.

Now sett= 1/qand infer that

ln

(

bq

q

+

ap

p

)

≥lnb+ lna= ln(ab).

Exponentiating both sides yields the desired result, namely that

bq

q

+

ap

p

≥ab.

Theorem 2.(H ̈older’s Inequality)Given real numbersx 1 ,...,xn, y 1 ,...,yn,

and given non-negative real numberspandqsuch that

1

p

+

1

q

= 1then

∑n

i=1

|xiyi|≤

Ñn

∑

i=1

|xi|p

é 1 /pÑ n

∑

j=1

|yj|q

é 1 /q

Proof. Let

A=

Ñn

∑

i=1

|xi|p

é 1 /p

, B=

Ñn

∑

i=1

|yi|q

é 1 /q

(^5) Another way to say this is, of course, that the graph ofy= lnxisconcave down.