If \(a,b,c \in \mathbb{R}^{+}\) , such that \(abc=1\) , then prove that \(\displaystyle\sum_{cyc} \dfrac{a}{b} \geq \sum_{cyc} a\)

**Instruction:** Those who have complete solution ready , please don't post it immediately.Post it minimum after 3 days of posting this note or give a small hint but not the solution.

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## Comments

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TopNewestMultiplying by \(abc\) and making the right side homogenous, it suffices to prove: \[a^2c + b^2a + c^2b \geq (abc)^{\frac{2}{3}}(a+b+c) = a^{\frac{5}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}} + a^{\frac{2}{3}}b^{\frac{5}{3}}c^{\frac{2}{3}} + a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{5}{3}} \] By AM-GM you know: \[a^2c + a^2c + ab^2 \geq 3a^{\frac{5}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}\] \[b^2a + b^2a + bc^2 \geq 3a^{\frac{2}{3}}b^{\frac{5}{3}}c^{\frac{2}{3}}\] \[c^2b + c^2b + ca^2 \geq 3a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{5}{3}}\] Add them and you are done.

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How and why did you develop such approach....It's not obvious at all.....Plz. provide some better suggestions...

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Apply AM GM(twice)

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what about the inequality posted by me ? it's hard try to solve it.

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I tried it but I didn't get it. I think its too tough for RMO level. You can post a solution there (if you have).

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Can I post my solutions now? I used MG-MA (reversed).

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You can post it now :)

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Post it after one or two days.

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i think we can make subsitutions... a = x/y, b = y/z and c = z/x

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I don't think such a substitution is not allowed. Take the example of 1/2 , 3/4 , 8/3. This is not an example like a=x/y,b=y/z,c=z/x.

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Also , if in your method , you are stuck some where , you are welcome to show your working.

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I recommend you to solve it completely and then post it.If you have some thoughts/ideas , you can post them after a complete solution is posted.

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Does \(\displaystyle\sum_{cyc} \dfrac{a}{b}\) mean \(\dfrac{a}{b} +\dfrac{b}{c} + \dfrac{c}{a} \) or \(\dfrac{a}{b} +\dfrac{b}{c} + \dfrac{c}{a} +\dfrac{b}{a} +\dfrac{c}{b} + \dfrac{a}{c} \) ??

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This can be proved using AM-GM only

\(\large{\frac { a }{ b } +\frac { b }{ c } +\frac { c }{ a } \ge 3\sqrt { \frac { a }{ b } \times \frac { b }{ c } \times \frac { c }{ a } } \\ \frac { a }{ b } +\frac { b }{ c } +\frac { c }{ a } \ge 3\\ \\ a+b+c\ge 3\sqrt { abc } \\ a+b+c\ge 3}\)

Dividing the new equation you will get the inequality mentioned in the question or reverse. But reverse is not acceptable because any great variation in \(a,b,c\) will not satisfy it. Sorry Nihar for posting the full solution.

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I don't think you can divide the inequalities. For example (exaggerated), \[ 2 \geq 1\] \[ 200000000000000 \geq 1 \] Dividing the first inequality by the second yields an incorrect inequality :\ Check out my solution.

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