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Algebra Difficulty 4.9 AIME Prove it China

Given real numbers aa, bb, cc, satisfying a+b+c=1a + b + c = 1, prove that
10(a3+b3+c3)−9(a5+b5+c5)≥1.10(a^3 + b^3 + c^3) - 9(a^5 + b^5 + c^5) \ge 1. (posed by Li Shenghong)

Solution

Since ∑a3=1−3Π(a+b)\sum a^3 = 1 - 3\Pi(a+b), ∑a5=1−5Π(a+b)[∑a2+∑ab]\sum a^5 = 1 - 5\Pi(a+b)[\sum a^2 + \sum ab], therefore, the original inequality holds
⇔10[1−3Π(a+b)]−9[1−5Π(a+b)(∑a2+∑ab)]≥1⇔45Π(a+b)(∑a2+∑ab)≥30Π(a+b)⇔3(∑a2+∑ab)≥2=2(∑a)2=2(∑a2+2∑ab)⇔∑a2≥∑ab. \begin{align*} &\Leftrightarrow 10[1 - 3\Pi(a+b)] - 9[1 - 5\Pi(a+b)(\sum a^2 + \sum ab)] \ge 1 \\ &\Leftrightarrow 45\Pi(a+b)(\sum a^2 + \sum ab) \ge 30\Pi(a+b) \\ &\Leftrightarrow 3(\sum a^2 + \sum ab) \ge 2 = 2(\sum a)^2 = 2(\sum a^2 + 2\sum ab) \\ &\Leftrightarrow \sum a^2 \ge \sum ab. \end{align*}
From a2+b2≥2aba^2 + b^2 \ge 2ab, b2+c2≥2bcb^2 + c^2 \ge 2bc and c2+a2≥2acc^2 + a^2 \ge 2ac, we have 2∑a2≥2∑ab2\sum a^2 \ge 2\sum ab, i.e., ∑a2≥∑ab\sum a^2 \ge \sum ab. Therefore the original inequality holds.

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