51st Ukrainian National Mathematical Olympiad, 3rd Round

Question

For nonnegative real numbers a, b, c, with the sum that does not exceed 2, prove ab(a^2 + b^2) + bc(b^2 + c^2) + ca(c^2 + a^2) 2.

Accepted Answer

It is enough to prove the inequality for

a + b + c = 2

. Let us denote

x = ab + b c + a c

, then

a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2 (ab + b c + a c) = 4 - 2 (ab + b c + a c) = 2 (2 - x) .

From the obvious inequality

x (2 - x) \leq 1

, we get

2 + 2 ab c \geq 2 \geq 2 x (2 - x) = (ab + b c + a c) (a^{2} + b^{2} + c^{2}) = = ab (a^{2} + b^{2}) + c b (c^{2} + b^{2}) + a c (a^{2} + c^{2}) + ab c (a + b + c),

Using the equality

2 ab c = ab c (a + b + c)

, we arrive at the conclusion.

Equality occurs, for example,

a = b = 1

,

c = 0

.