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\[\large \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq\frac{4}{3}\bigg(\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}\bigg)\] 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?

Note: Don't comment \(a=b=c\), it's really irresponsible.

Note by Gurīdo Cuong
7 months ago

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@Svatejas Shivakumar Oops, I believe you've made an error. This inequality is not trivial, because, you showed,
\(\dfrac{a^2}{a^2+bc}+\dfrac{b^2}{b^2+ac}+\dfrac{c^2}{c^2+ab} \ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca} \\ \) 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 · 6 months, 3 weeks ago

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@Ameya Daigavane Yes you are right. Do you know how to proceed with this problem? Svatejas Shivakumar · 6 months, 3 weeks ago

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@Svatejas Shivakumar I'm working on it, with some normalisation technique. Ameya Daigavane · 6 months, 3 weeks ago

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I proved it up to the lhs is greater than or equal to 427abc/( (a+b+c)^3+27abc) Lakshya Sinha · 7 months ago

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@Lakshya Sinha can you show me how? Gurīdo Cuong · 7 months ago

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@Gurīdo Cuong If you are on Slack, I can send you there. Lakshya Sinha · 7 months ago

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@Lakshya Sinha yes I am, can you send it to me? Gurīdo Cuong · 7 months ago

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@Gurīdo Cuong I am busy with my finals so I would send you it later make sure you check it Lakshya Sinha · 7 months ago

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@Lakshya Sinha alright, good luck dude Gurīdo Cuong · 7 months ago

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@Gurīdo Cuong Thanks, but please check it out later, but it may contain flaws and feel free to spot the mistakes Lakshya Sinha · 7 months ago

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How did you proved the first part Lakshya Sinha · 7 months ago

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@Lakshya Sinha First can be proved easily using AM-GM inequality. Harsh Shrivastava · 7 months ago

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@Harsh Shrivastava Yes just saw that Lakshya Sinha · 7 months ago

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