Estonian Mathematical Olympiad

Question

Find all functions f from the set of all non-negative real numbers to the set of all real numbers such that f(1) = 1 and (f(x+y))^2 f(x^2 - 2xy + y^2) for all real numbers x and y that satisfy the inequality x + y 0.

Accepted Answer

The given inequality can be rewritten as

(f (x + y))^{2} \leq f ((x - y)^{2}) .

Take

y = 1 - x

. Then

1 = 1^{2} = (f (1))^{2} \leq f ((2 x - 1)^{2})

. As

(2 x - 1)^{2}

obtains all non-negative real values, we can conclude that

1 \leq f (z)

for every non-negative real number

z

. Now take

x \geq - \frac{1}{2}

and

y = 1 + x

. Then

(f (2 x + 1))^{2} \leq f (1) = 1

. This implies

f (2 x + 1) \leq 1

because all values of

f

are non-negative by the above. As

2 x + 1

obtains all non-negative real values, we can deduce that

f (z) \leq 1

for every non-negative real number

z

.

Consequently,

f (z) = 1

for every non-negative real number

z

. This function satisfies all conditions of the problem.