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RMO Inequality Practice

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.

Note by Nihar Mahajan
1 year, 8 months ago

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Multiplying 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. Alan Yan · 1 year, 8 months ago

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what about the inequality posted by me ? it's hard try to solve it. Shivam Jadhav · 1 year, 8 months ago

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@Shivam Jadhav 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). Nihar Mahajan · 1 year, 8 months ago

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Can I post my solutions now? I used MG-MA (reversed). Johanz Piedad · 1 year, 8 months ago

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@Johanz Piedad You can post it now :) Nihar Mahajan · 1 year, 8 months ago

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@Johanz Piedad Post it after one or two days. Nihar Mahajan · 1 year, 8 months ago

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i think we can make subsitutions... a = x/y, b = y/z and c = z/x Dev Sharma · 1 year, 8 months ago

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@Dev Sharma 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. Shrihari B · 1 year, 5 months ago

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@Dev Sharma Also , if in your method , you are stuck some where , you are welcome to show your working. Nihar Mahajan · 1 year, 8 months ago

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@Dev Sharma 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. Nihar Mahajan · 1 year, 8 months ago

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Does \(\displaystyle\sum_{cyc} \dfrac{a}{b}\) mean \(\dfrac{a}{b} +\dfrac{b}{c} + \dfrac{c}{a} \) or \(\dfrac{a}{b} +\dfrac{b}{c} + \dfrac{c}{a} +\dfrac{b}{a} +\dfrac{c}{b} + \dfrac{a}{c} \) ?? Yash Mehan · 9 months, 2 weeks ago

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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. Lakshya Sinha · 1 year, 8 months ago

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@Lakshya Sinha 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. Alan Yan · 1 year, 8 months ago

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