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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestApply AM GM(twice)

Log in to reply

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

Log in to reply

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.

Log in to reply

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.

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

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).

Log in to reply

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

Log in to reply

You can post it now :)

Log in to reply

Post it after one or two days.

Log in to reply

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

Log in to reply

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.

Log in to reply

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

Log in to reply

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.

Log in to reply