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Algebra Difficulty 5.7 AIME, harder Prove it Croatia

Let aa, bb and cc be positive real numbers such that a+b+c=2a + b + c = 2. Prove that
(a−1)2b+(b−1)2c+(c−1)2a≥14(a2+b2a+b+b2+c2b+c+c2+a2c+a). \frac{(a-1)^2}{b} + \frac{(b-1)^2}{c} + \frac{(c-1)^2}{a} \ge \frac{1}{4} \left( \frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c} + \frac{c^2+a^2}{c+a} \right).

Solution

By the Cauchy-Bunyakovsky-Schwarz inequality, we have
(a−1)2b+(b−1)2c≥(2−a−b)2b+c=c2b+c, \frac{(a-1)^2}{b} + \frac{(b-1)^2}{c} \ge \frac{(2-a-b)^2}{b+c} = \frac{c^2}{b+c},
and similar inequalities hold for all pairs. By adding the inequalities, we get
(a−1)2b+(b−1)2c+(c−1)2a≥12(b2a+b+c2b+c+a2c+a). \frac{(a-1)^2}{b} + \frac{(b-1)^2}{c} + \frac{(c-1)^2}{a} \ge \frac{1}{2} \left( \frac{b^2}{a+b} + \frac{c^2}{b+c} + \frac{a^2}{c+a} \right).
Note that the initial claim follows from this, because
b2a+b+c2b+c+a2c+a=a2a+b+b2b+c+c2c+a, \frac{b^2}{a+b} + \frac{c^2}{b+c} + \frac{a^2}{c+a} = \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+a},
which holds since
b2a+b+c2b+c+a2c+a−a2a+b−b2b+c−c2c+a=b2−a2a+b+c2−b2b+c+a2−c2c+a=(b−a)+(c−b)+(a−c)=0. \begin{aligned} \frac{b^2}{a+b} + \frac{c^2}{b+c} + \frac{a^2}{c+a} - \frac{a^2}{a+b} - \frac{b^2}{b+c} - \frac{c^2}{c+a} &= \frac{b^2-a^2}{a+b} + \frac{c^2-b^2}{b+c} + \frac{a^2-c^2}{c+a} \\ &= (b-a) + (c-b) + (a-c) = 0. \end{aligned}

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