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Algebra Difficulty 5.9 AIME, harder Prove it

Example 1.2.7 Let a,b,c>0,a+b+c=3a, b, c>0, a+b+c=3, prove: 11+2b2c+11+2c2a+11+2a2b≥1\frac{1}{1+2 b^{2} c}+\frac{1}{1+2 c^{2} a}+\frac{1}{1+2 a^{2} b} \geq 1

Solution

Prove: We use the following estimation
11+2b2c=1−2b2c1+2b2c≥1−2b2c33≥1−2(2b+c)9\frac{1}{1+2 b^{2} c}=1-\frac{2 b^{2} c}{1+2 b^{2} c} \geq 1-\frac{2 \sqrt[3]{b^{2} c}}{3} \geq 1-\frac{2(2 b+c)}{9}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.