51st Ukrainian National Mathematical Olympiad, 4th Round

Question

For any positive real numbers x, y, z with xyz=1 prove the following inequality: (-x+y+z)(x-y+z) + (x-y+z)(x+y-z) + (x+y-z)(-x+y+z) 3.

Accepted Answer

Without loss of generality, assume that

z = min {x, y, z}

. The inequality can be rewritten as

4 x y \leq 3 + (x + y - z)^{2} .

Since

x + y - z \geq 2 x y - z > 0

, we have that

(x + y - z)^{2} \geq (2 x y - z)^{2}

. So, it is sufficient to prove that

3 + (2 x y - z)^{2} \geq 4 x y

. From the problem condition

x y = \frac{1}{z}

, hence, we need to prove the following inequality

3 + (\frac{2}{z} - z)^{2} \geq \frac{4}{z} .

Expanding the brackets and multiplying by

z

, we obtain the inequality:

z^{2} + 3 \geq 4 z

, which follows from the AM-GM inequality:

z^{2} + 1 + 1 + 1 \geq 4 z^{2} \cdot 1 \cdot 1 \cdot 1 = 4 z .

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Alternative solution.

We can rewrite our inequality in the form

2 x y + 2 y z + 2 z x \leq 3 + x^{2} + y^{2} + z^{2} .

We will use the Schur's inequality

(u, v, w \geq 0)

:

u^{3} + v^{3} + w^{3} + 3 uv w \geq u^{2} v + v^{2} u + w^{2} u + w^{2} v + w^{2} u + w^{2} v + w^{2} u,

and the AM-GM inequality for two numbers. With

x = a^{3}

,

y = b^{3}

,

z = c^{3}

, we have $$

x^{2} + y^{2} + z^{2} + 3 = a^{6} + b^{6} + c^{6} + 3 a^{2} b^{2} c^{2} \geq 2 a^{3} b^{3} + 2 b^{3} c^{3} + 2 c^{3} a^{3} = 2 x y + 2 y z + 2 x z . \geq (a^{4} b^{2} + a^{2} b^{4}) + (b^{4} c^{2} + b^{2} c^{4}) + (c^{4} a^{2} + c^{2} a^{4})