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Algebra Difficulty 4.8 AIME Prove it Belarus

Prove that
a2a+b+b2b+c≥3a+2b−c4 \frac{a^2}{a+b} + \frac{b^2}{b+c} \ge \frac{3a+2b-c}{4}
for all positive real a,b,ca, b, c.

(D. Pirshtuk)

Solution

(2x−y)2≥0  ⟺  x2y≥4x−y4 (2x - y)^2 \ge 0 \iff \frac{x^2}{y} \ge \frac{4x - y}{4}
for any positive x,yx, y.

Therefore,
a2a+b+b2b+c≥4a−(a+b)4+4b−(b+c)4=3a+2b−c4, \frac{a^2}{a+b} + \frac{b^2}{b+c} \ge \frac{4a - (a+b)}{4} + \frac{4b - (b+c)}{4} = \frac{3a + 2b - c}{4},
as required.

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