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a/(a+b) x (a+2b)/(a+3b) < \sqrt{ a /(a+4b)}

Source: 2010 Moldova JBMO TST p5

February 25, 2021
inequalities

Problem Statement

For any strictly positive numbers aa and bb , prove the inequality aa+ba+2ba+3b<aa+4b.\frac{a}{a+b} \cdot \frac{a+2b}{a+3b} < \sqrt{ \frac{a}{a+4b}}.