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# Need help on this 2

$\large \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq\frac{4}{3}\bigg(\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}\bigg)$ 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
9 months, 1 week ago

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Comment deleted 9 months ago

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). · 9 months ago

Yes you are right. Do you know how to proceed with this problem? · 9 months ago

I'm working on it, with some normalisation technique. · 9 months ago

I proved it up to the lhs is greater than or equal to 427abc/( (a+b+c)^3+27abc) · 9 months, 1 week ago

can you show me how? · 9 months, 1 week ago

If you are on Slack, I can send you there. · 9 months, 1 week ago

yes I am, can you send it to me? · 9 months, 1 week ago

I am busy with my finals so I would send you it later make sure you check it · 9 months, 1 week ago

alright, good luck dude · 9 months, 1 week ago

Thanks, but please check it out later, but it may contain flaws and feel free to spot the mistakes · 9 months, 1 week ago

How did you proved the first part · 9 months, 1 week ago

First can be proved easily using AM-GM inequality. · 9 months, 1 week ago