×

# 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, 10 months ago

Sort by:

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. · 1 year, 10 months ago

what about the inequality posted by me ? it's hard try to solve it. · 1 year, 10 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). · 1 year, 10 months ago

Can I post my solutions now? I used MG-MA (reversed). · 1 year, 10 months ago

You can post it now :) · 1 year, 10 months ago

Post it after one or two days. · 1 year, 10 months ago

i think we can make subsitutions... a = x/y, b = y/z and c = z/x · 1 year, 10 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. · 1 year, 7 months ago

Also , if in your method , you are stuck some where , you are welcome to show your working. · 1 year, 10 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. · 1 year, 10 months ago

Does $$\displaystyle\sum_{cyc} \dfrac{a}{b}$$ mean $$\dfrac{a}{b} +\dfrac{b}{c} + \dfrac{c}{a}$$ or $$\dfrac{a}{b} +\dfrac{b}{c} + \dfrac{c}{a} +\dfrac{b}{a} +\dfrac{c}{b} + \dfrac{a}{c}$$ ?? · 11 months, 2 weeks 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. · 1 year, 10 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. · 1 year, 10 months ago