Advanced book on Mathematics Olympiad

524 Real Analysis

=

f(x 1 )+···+f(x 2 k)+f(x 2 k+ 1 )+···+f(x 2 k+ 1 )

2 k+^1

.

Next, we show that

f

(

x 1 +x 2 +···+xn

n

)

≤

f(x 1 )+f(x 2 )+···+f(xn)

n

, for allx 1 ,x 2 ...,xn.

Assuming that the inequality holds for anynpoints, we prove that it holds for anyn− 1
points as well. Consider the pointsx 1 ,x 2 ,...,xn− 1 and definexn=x^1 +x^2 +···+n− 1 xn−^1. Using
the induction hypothesis, we can write

f

(

x 1 +···+xn− 1 +x^1 +···+n− 1 xn−^1

n

)

≤

f(x 1 )+···+f(xn− 1 )+f

(x 1 +···+xn− 1

n− 1

)

n

.

This is the same as

f

(

x 1 +···+xn− 1

n− 1

)

≤

f(x 1 )+···+f(xn− 1 )

n

+

1

n

f

(

x 1 +···+xn− 1

n− 1

)

.

Moving the last term on the right to the other side gives the desired inequality. Starting

with a sufficiently large power of 2 we can cover the case of any positive integern.

In the inequality

f

(

x 1 +x 2 +···+xn

n

)

≤

f(x 1 )+f(x 2 )+···+f(xn)

n

that we just proved, for somem<nsetx 1 =x 2 = ··· =xm=xandxm+ 1 =xm+ 2 =

··· =xn=y. Then

f

(m

n

x+

(

1 −

m

n

)

y

)

≤

m

n

f(x)+

(

1 −

m

n

)

f(y).

Becausefis continuous we can pass to the limit withmn→λto obtain the desired

f (λx+( 1 −λ)y)≤λf (x)+( 1 −λ)f (y),

which characterizes convex functions.

430.First solution: Fixn≥1. For each integeri, define

(^) i=f

(

i+ 1

n

)

−f

(

i

n

)

.

If in the inequality from the statement we substitutex=i+n^2 andy=in, we obtain