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

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TopNewestComment deleted Mar 06, 2016

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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).

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

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

27abc/( (a+b+c)^3+27abc)Log in to reply

can you show me how?

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If you are on Slack, I can send you there.

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How did you proved the first part

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First can be proved easily using AM-GM inequality.

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Yes just saw that

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