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Algebra Difficulty 5.0 AIME, harder Prove it Romania

Let a,b,c,d≥0a, b, c, d \ge 0 be real numbers such that a+b+c+d=1a + b + c + d = 1. Prove that
a+(b−c)26+(c−d)26+(d−b)26+b+c+d≤2. \sqrt{a + \frac{(b-c)^2}{6} + \frac{(c-d)^2}{6} + \frac{(d-b)^2}{6}} + \sqrt{b} + \sqrt{c} + \sqrt{d} \le 2.

Solution

Observe that (b−c)2≤2(b−c)2(b-c)^2 \le 2(\sqrt{b} - \sqrt{c})^2, since (b+c)2=b+c+2bc≤2b+2c≤2(\sqrt{b} + \sqrt{c})^2 = b + c + 2\sqrt{bc} \le 2b + 2c \le 2. Thus, a+∑(b−c)26≤a+13∑(b−c)2=1−13(∑b)2a + \sum \frac{(b-c)^2}{6} \le a + \frac{1}{3} \sum (\sqrt{b} - \sqrt{c})^2 = 1 - \frac{1}{3}(\sum \sqrt{b})^2.

To end the proof, it suffices to show that S+1−S2/3≤2S + \sqrt{1 - S^2/3} \le 2, where S=b+c+dS = \sqrt{b} + \sqrt{c} + \sqrt{d}.

Since S≤2S \le 2, this is equivalent to 1−S2/3≤4−4S+S21 - S^2/3 \le 4 - 4S + S^2, which rewrites as (2S−3)2≥0(2S - 3)^2 \ge 0, obviously true.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.