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Algebra Difficulty 4.6 AIME Prove it Soviet Union

Problem:
Show that {x}4>{x}1/2\{x\}^4 > \{x\} - 1/2 for all real xx.

Solution

Solution:
{x}4{x}+1/2=({x}21/2)2+({x}1/2)20\{x\}^4 - \{x\} + 1/2 = (\{x\}^2 - 1/2)^2 + (\{x\} - 1/2)^2 \geq 0. We could only have equality if {x}2={x}=1/2\{x\}^2 = \{x\} = 1/2, which is impossible, so the inequality is strict.

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