Balti Tee 2011 · Valikvooru ülesanne

Algebra

Let $x$ , $y$ , $z$ , $t$ be positive real numbers such that $xyzt = 1$ and

\frac{x}{y} + \frac{y}{z} + \frac{z}{t} + \frac{t}{x} \leq x + y + z + t.

Prove that

\frac{y}{x} + \frac{z}{y} + \frac{t}{z} + \frac{x}{t} \geq x + y + z + t.

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Lahendus

By the arithmetic mean-geometric mean inequality we have

x = \sqrt[4]{x^4} = \sqrt[4]{\frac{x^4}{xyzt}} = \sqrt[4]{\frac{x^3}{yzt}} = \sqrt[4]{\frac{x}{y} \cdot \frac{x}{t} \cdot \frac{t}{z} \cdot \frac{x}{t}} \leq \frac{1}{4} \left( \frac{x}{y} + \frac{x}{t} + \frac{t}{z} + \frac{x}{t} \right) = \frac{1}{4} \left( \frac{x}{y} + 2 \cdot \frac{x}{t} + \frac{t}{z} \right).

Similarly we show that

y \leq \frac{1}{4} \left( \frac{y}{z} + 2 \cdot \frac{y}{x} + \frac{x}{t} \right), \quad z \leq \frac{1}{4} \left( \frac{z}{t} + 2 \cdot \frac{z}{y} + \frac{y}{x} \right), \quad t \leq \frac{1}{4} \left( \frac{t}{x} + 2 \cdot \frac{t}{z} + \frac{z}{y} \right).

Adding together the four inequalities and applying the assumed inequality we obtain

x+y+z+t \leq \frac{1}{4} \left( \frac{x}{y} + \frac{y}{z} + \frac{z}{t} + \frac{t}{x} \right) + \frac{3}{4} \left( \frac{y}{x} + \frac{z}{y} + \frac{t}{z} + \frac{x}{t} \right) \leq \frac{1}{4}(x+y+z+t) + \frac{3}{4} \left( \frac{y}{x} + \frac{z}{y} + \frac{t}{z} + \frac{x}{t} \right).

The assertion of the problem follows immediately.