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Algebra Difficulty 5.2 AIME, harder Prove it Ukraine

Show that for positive numbers x,y,z,tx, y, z, t the following inequality holds:
x8+1x4+y8+1y4+z8+1z4+t8+1t42(xy+yz+zt+tx). \frac{x^8+1}{x^4} + \frac{y^8+1}{y^4} + \frac{z^8+1}{z^4} + \frac{t^8+1}{t^4} \ge 2 \cdot \left( \frac{x}{y} + \frac{y}{z} + \frac{z}{t} + \frac{t}{x} \right).

Solution

The inequality can be rewritten as follows:
(x4+1z4+z4+1y4)+(y4+1t4+t4+1z4)+(z4+1x4+x4+1t4)+(t4+1y4+y4+1x4)4(xy+yz+zt+tx). \begin{aligned} & (x^4 + \frac{1}{z^4} + z^4 + \frac{1}{y^4}) + (y^4 + \frac{1}{t^4} + t^4 + \frac{1}{z^4}) + (z^4 + \frac{1}{x^4} + x^4 + \frac{1}{t^4}) + (t^4 + \frac{1}{y^4} + y^4 + \frac{1}{x^4}) \\ & \ge 4 \cdot \left( \frac{x}{y} + \frac{y}{z} + \frac{z}{t} + \frac{t}{x} \right). \end{aligned}

(x4+1z4)+(z4+1y4)2x41z4+2z41y4=2x2z2+2z2y24x2z2z2y2=4xyz,(x^4 + \frac{1}{z^4}) + (z^4 + \frac{1}{y^4}) \geq 2\sqrt{x^4 \cdot \frac{1}{z^4}} + 2\sqrt{z^4 \cdot \frac{1}{y^4}} = \frac{2x^2}{z^2} + \frac{2z^2}{y^2} \geq 4\sqrt{\frac{x^2}{z^2} \cdot \frac{z^2}{y^2}} = \frac{4x}{y z},
that finishes the proof.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.