Maths Olympiad Prep

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, 2011

Algebra Difficulty 7.7 National Olympiad, round 2 Prove it Baltic Way

Let xx, yy, zz, tt be positive real numbers such that xyzt=1xyzt = 1 and
xy+yz+zt+tx≤x+y+z+t. \frac{x}{y} + \frac{y}{z} + \frac{z}{t} + \frac{t}{x} \leq x + y + z + t.
Prove that
yx+zy+tz+xt≥x+y+z+t. \frac{y}{x} + \frac{z}{y} + \frac{t}{z} + \frac{x}{t} \geq x + y + z + t.

Solution

By the arithmetic mean-geometric mean inequality we have
x=x44=x4xyzt4=x3yzt4=xy⋅xt⋅tz⋅xt4≤14(xy+xt+tz+xt)=14(xy+2⋅xt+tz). 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≤14(yz+2⋅yx+xt),z≤14(zt+2⋅zy+yx),t≤14(tx+2⋅tz+zy). 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≤14(xy+yz+zt+tx)+34(yx+zy+tz+xt)≤14(x+y+z+t)+34(yx+zy+tz+xt). 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.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.