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.

## Comments

Sort by:

TopNewestMultiplying 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, 4 months ago

Log in to reply

what about the inequality posted by me ? it's hard try to solve it. – Shivam Jadhav · 1 year, 4 months ago

Log in to reply

– Nihar Mahajan · 1 year, 4 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).Log in to reply

Can I post my solutions now? I used MG-MA (reversed). – Johanz Piedad · 1 year, 4 months ago

Log in to reply

– Nihar Mahajan · 1 year, 3 months ago

You can post it now :)Log in to reply

– Nihar Mahajan · 1 year, 4 months ago

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 – Dev Sharma · 1 year, 4 months ago

Log in to reply

– Shrihari B · 1 year, 1 month 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.Log in to reply

– Nihar Mahajan · 1 year, 4 months ago

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

– Nihar Mahajan · 1 year, 4 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.Log in to reply

Does \(\displaystyle\sum_{cyc} \dfrac{a}{b}\) mean \(\dfrac{a}{b} +\dfrac{b}{c} + \dfrac{c}{a} \) or \(\dfrac{a}{b} +\dfrac{b}{c} + \dfrac{c}{a} +\dfrac{b}{a} +\dfrac{c}{b} + \dfrac{a}{c} \) ?? – Yash Mehan · 5 months, 1 week ago

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. – Lakshya Sinha · 1 year, 3 months ago

Log in to reply

– Alan Yan · 1 year, 3 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.Log in to reply