Austrian Mathematical Olympiad

Question

Let a, b, c be positive real numbers with a + b + c = 1. Prove that a2a+1 + b3b+1 + c6c+1 12. When does equality hold?

Accepted Answer

We will use the following inequalities:

\frac{a}{2 a + 1} \leq \frac{2 a + 1}{8}, \frac{b}{3 b + 1} \leq \frac{3 b + 1}{12} and \frac{c}{6 c + 1} \leq \frac{6 c + 1}{24} .

They are an immediate consequence of the arithmetic-geometric mean inequality with the pairs of values

1

and

2 a

,

1

and

3 b

, and

1

and

6 c

, so that equality holds for

a = 1/2

,

b = 1/3

and

c = 1/6

. From these inequalities and the condition

a + b + c = 1

, we obtain

\frac{a}{2 a + 1} + \frac{b}{3 b + 1} + \frac{c}{6 c + 1} \leq \frac{2 a + 1}{8} + \frac{3 b + 1}{12} + \frac{6 c + 1}{24} = \frac{1}{2} .

Therefore, the given inequality is true and equality holds exactly for

a = 1/2

,

b = 1/3

and

c = 1/6

.