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Algebra Difficulty 6.1 National olympiad Prove it

9・152 Let a,b,ca, b, c all be greater than 1, prove:
2(log⁡baa+b+log⁡bbb+c+log⁡acc+a)⩾9a+b+c.2\left(\frac{\log _{b} a}{a+b}+\frac{\log _{b} b}{b+c}+\frac{\log _{a} c}{c+a}\right) \geqslant \frac{9}{a+b+c} .

Solution

[Proof] Since a>1,b>1,c>1a>1, b>1, c>1, we have
log⁡ba⋅log⁡cb⋅log⁡ac=1\log _{b} a \cdot \log _{c} b \cdot \log _{a} c=1

Using the AM-GM inequality, we get
log⁡baa+b+log⁡bbb+c+log⁡acc+a⩾31(a+b)(b+c)(c+a)3\frac{\log _{b} a}{a+b}+\frac{\log _{b} b}{b+c}+\frac{\log _{a} c}{c+a} \geqslant 3 \sqrt[3]{\frac{1}{(a+b)(b+c)(c+a)}}

Using the AM-GM inequality again, we obtain
31(a+b)(b+c)(c+a)3⩾92(a+b+c)3 \sqrt[3]{\frac{1}{(a+b)(b+c)(c+a)}} \geqslant \frac{9}{2(a+b+c)}

Thus, the inequality to be proven holds.

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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.