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sum (a^2-b^2)/c >= 3a - 4b + c if a >= b >= c > 0

Source: 2000 Moldova JBMO TST p5

February 20, 2021
algebrainequalities

Problem Statement

Let the real numbers a,b,ca, b, c be such that a≥b≥c>0a \ge b \ge c > 0. Show that a2−b2c+c2−b2a+a2−c2b≥3a−4b+c.\frac{a^2-b^2}{c}+ \frac{c^2-b^2}{a}+ \frac{a^2-c^2}{b}\ge 3a - 4b + c. When does equality hold?
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