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.