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Algebra Difficulty 4.6 AIME Prove it Romania

If aa, bb, cc are positive real numbers, prove that
a(a+2b)3+b(b+2c)3+c(c+2a)3≥1a+b+c. \frac{a}{\sqrt{(a+2b)^3}} + \frac{b}{\sqrt{(b+2c)^3}} + \frac{c}{\sqrt{(c+2a)^3}} \ge \frac{1}{\sqrt{a+b+c}}.

Solution

From Hölder's inequality we have:
∑cycla(a+2b)3⋅∑cyclaa+2b⋅∑cyclaa+2b⋅∑cyclaa+2b≥(a+b+c)4, \sum_{cycl} \frac{a}{\sqrt{(a+2b)^3}} \cdot \sum_{cycl} a\sqrt{a+2b} \cdot \sum_{cycl} a\sqrt{a+2b} \cdot \sum_{cycl} a\sqrt{a+2b} \ge (a+b+c)^4,
and from Cauchy-Buniakowsky-Schwarz it follows that
(∑cyclaa+2b)2=(∑cycla⋅a(a+2b))2≤(a+b+c)⋅∑cycl(a2+2ab)=(a+b+c)3 \left(\sum_{cycl} a\sqrt{a+2b}\right)^2 = \left(\sum_{cycl} \sqrt{a \cdot \sqrt{a(a+2b)}}\right)^2 \le (a+b+c) \cdot \sum_{cycl} (a^2+2ab) \\ = (a+b+c)^3

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