\[\large \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq\frac{4}{3}\bigg(\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}\bigg)\] If \(a,b\) and \(c\) are positive reals, prove the inequality above.

I've proven \[\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq 2\] So now I have to prove \[\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}\leq\frac{3}{2}\] I've tried to prove it but then realised that what I was trying to prove is wrong, can somebody help me?

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\(\dfrac{a^2}{a^2+bc}+\dfrac{b^2}{b^2+ac}+\dfrac{c^2}{c^2+ab} \ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca} \\ \) and \(\dfrac{3}{2} \ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca}\)

But this does not imply \(\dfrac{3}{2} \ge \dfrac{a^2}{a^2+bc}+\dfrac{b^2}{b^2+ac}+\dfrac{c^2}{c^2+ab} \) which is not true for all (a, b, c). – Ameya Daigavane · 10 months, 3 weeks ago

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– Svatejas Shivakumar · 10 months, 3 weeks ago

Yes you are right. Do you know how to proceed with this problem?Log in to reply

– Ameya Daigavane · 10 months, 3 weeks ago

I'm working on it, with some normalisation technique.Log in to reply

I proved it up to the lhs is greater than or equal to 4

27abc/( (a+b+c)^3+27abc) – Lakshya Sinha · 11 months agoLog in to reply

– Gurīdo Cuong · 11 months ago

can you show me how?Log in to reply

Slack, I can send you there. – Lakshya Sinha · 11 months ago

If you are onLog in to reply

– Gurīdo Cuong · 11 months ago

yes I am, can you send it to me?Log in to reply

– Lakshya Sinha · 11 months ago

I am busy with my finals so I would send you it later make sure you check itLog in to reply

– Gurīdo Cuong · 11 months ago

alright, good luck dudeLog in to reply

– Lakshya Sinha · 11 months ago

Thanks, but please check it out later, but it may contain flaws and feel free to spot the mistakesLog in to reply

How did you proved the first part – Lakshya Sinha · 11 months ago

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– Harsh Shrivastava · 11 months ago

First can be proved easily using AM-GM inequality.Log in to reply

– Lakshya Sinha · 11 months ago

Yes just saw thatLog in to reply