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Algebra Difficulty 4.5 AIME Prove it Croatia

Let aa, bb and cc be positive real numbers. Prove the inequality
a2a+b+b2b+c≥3a+2b−c4. \frac{a^2}{a+b} + \frac{b^2}{b+c} \ge \frac{3a+2b-c}{4}.
(Belarus 2010)

Solution

a2a+b+a+b4≥aandb2b+c+b+c4≥b. \frac{a^2}{a+b} + \frac{a+b}{4} \ge a \quad \text{and} \quad \frac{b^2}{b+c} + \frac{b+c}{4} \ge b.
Adding these two inequalities we get
(a2a+b+a+b4)+(b2b+c+b+c4)≥a+b=3a+2b−c4+a+b4+b+c4. \left( \frac{a^2}{a+b} + \frac{a+b}{4} \right) + \left( \frac{b^2}{b+c} + \frac{b+c}{4} \right) \ge a+b = \frac{3a+2b-c}{4} + \frac{a+b}{4} + \frac{b+c}{4}.
If we subtract a+b4+b+c4\frac{a+b}{4} + \frac{b+c}{4} from both sides, the result follows.

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