Advanced book on Mathematics Olympiad

Real Analysis 523

or

AsinB−AsinC−CsinB≥BsinA−CsinA−BsinC.

Moving the negative terms to the other side and substituting the sides of the triangle for
the sines, we obtain the inequality from the statement.

428.Fixx 0 ∈(a, b)and letαandβbe two limit points off:αfrom the left andβfrom
the right. We want to prove that they are equal. If not, without loss of generality we can
assumeα<β. We argue from Figure 68. Choosex<x 0 andy>x 0 very close tox 0
such that|f(x)−α|and|f(y)−β|are both very small. Becauseβis a limit point off
atx 0 , there will exist points on the graph offclose to(x 0 ,β), hence above the segment
joining(x, f (x))and(y, f (y)). But this contradicts the convexity off. Henceα=β.
Because all limit points from the left are equal to all limit points from the right,f
has a limit atx 0. Now redo the above argument forx=x 0 to conclude that the limit is
equal to the value of the function atx 0. Hencefis continuous atx 0.

Figure 68

429.The key point of the solution is Cauchy’s method of backward induction discussed
in the first chapter of the book. We first prove that for any positive integerkand points
x 1 ,x 2 ,...,x 2 k, we have

f

(

x 1 +x 2 +···+x 2 k

2 k

)

≤

f(x 1 )+f(x 2 )+···+f(x 2 k)

2 k

.

The base case is contained in the statement of the problem, while the inductive step is

f

(

x 1 +···+x 2 k+x 2 k+ 1 +···+x 2 k+ 1

2 k+^1

)

≤

f

(x

1 +···+x 2 k

2 k

)

+f

(x

2 k+ 1 +···+x 2 k+^1

2 k

)

2

≤

f(x 1 )+···+f(x 2 k)

2 k +

f(x 2 k+ 1 )+···+f(x 2 k+ 1 )

2 k

2