# 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
4 years, 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.

- 4 years, 8 months ago

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

- 2 years, 8 months ago

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

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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.

- 4 years, 6 months ago

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

- 4 years, 8 months ago

Post it after one or two days.

- 4 years, 8 months ago

You can post it now :)

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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).

- 4 years, 8 months ago

Apply AM GM(twice)

- 1 year, 8 months ago

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

- 3 years, 10 months ago

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.

- 4 years, 8 months ago

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.

- 4 years, 8 months ago