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Geometry Difficulty 4.9 AIME Prove it Ukraine

For which real numbers x>1x > 1 there exists a triangle with sides of lengths x4+x3+2x2+x+1x^4 + x^3 + 2x^2 + x + 1, 2x3+x2+2x+12x^3 + x^2 + 2x + 1 and x41x^4 - 1?

Solution

Дійсно, при всіх x>1x > 1, як нескладно переконатися, мають місце нерівності
x4+x3+2x2+x+1>2x3+x2+2x+1, x^4 + x^3 + 2x^2 + x + 1 > 2x^3 + x^2 + 2x + 1,
x4+x3+2x2+x+1>x41, x^4 + x^3 + 2x^2 + x + 1 > x^4 - 1,
x4+x3+2x2+x+1<(x41)+(2x3+x2+2x+1). x^4 + x^3 + 2x^2 + x + 1 < (x^4 - 1) + (2x^3 + x^2 + 2x + 1).

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