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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
2 years 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 - 2 years ago

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

Anubhav Mahapatra - 1 week, 4 days ago

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

Shivam Jadhav - 2 years ago

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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).

Nihar Mahajan - 2 years ago

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

Johanz Piedad - 2 years ago

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You can post it now :)

Nihar Mahajan - 2 years ago

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Post it after one or two days.

Nihar Mahajan - 2 years ago

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

Dev Sharma - 2 years ago

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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.

Shrihari B - 1 year, 10 months ago

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

Nihar Mahajan - 2 years ago

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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.

Nihar Mahajan - 2 years 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 - 1 year, 2 months 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 - 2 years ago

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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.

Alan Yan - 2 years ago

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