Advanced book on Mathematics Olympiad

3.2 Continuity, Derivatives, and Integrals 143

∑n

i= 1

xiyi≤

(n

∑

i= 1

xip

) 1 /p( n

∑

i= 1

yqi

) 1 /q

,

with equality if and only if the two vectors(x 1 ,x 2 ,...,xn)and(y 1 ,y 2 ,...,yn)are
parallel.

Proof.The second derivative off:( 0 ,∞)→R,f(x)=lnx,isf′′(x)=−x^12 , which
is negative. So this function is concave. Settingλ=p^1 , we obtain

lnX^1 /pY^1 /q=

1

p

lnX+

1

q

lnY≤ln

(

1

p

X+

1

q

Y

)

, for allX, Y > 0 ;

hence

X^1 /pY^1 /q≤

1

p

X+

1

q

Y.

Using this fact, if we letX=

∑

ix

p

i andY=

∑

iy

q

i, then

1

X^1 /pY^1 /q

∑n

i= 1

xiyi=

∑n

i= 1

(

xip

X

) 1 /p(

yiq

Y

) 1 /q

≤

∑n

i= 1

(

1

p

·

xip

X

+

1

q

·

yiq

Y

)

=

(

1

p

+

1

q

)

= 1.

Hence

∑n

i= 1

xiyi≤X^1 /pY^1 /q=

(n

∑

i= 1

xip

) 1 /p(n

∑

i= 1

yqi

) 1 /q

,

and the inequality is proved. 

By analogy, a sequence(an)n≥ 0 is called convex if

an≤

an+ 1 +an− 1

2

, for alln≥ 1 ,

and concave if(−an)nis convex. Equivalently, a sequence is convex if its second
difference (derivative) is nonnegative, and concave if its second difference is nonpositive.
The following example motivates why convex sequences and functions should be studied
together.

Example.Let(an)nbe a bounded convex sequence. Prove that

lim

n→∞

(an+ 1 −an)= 0.