# 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 by P C
2 years, 9 months ago

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Comment deleted Mar 06, 2016

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).

- 2 years, 9 months ago

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

- 2 years, 9 months ago

I'm working on it, with some normalisation technique.

- 2 years, 9 months ago

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

- 2 years, 9 months ago

can you show me how?

- 2 years, 9 months ago

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

- 2 years, 9 months ago

yes I am, can you send it to me?

- 2 years, 9 months ago

@P C I am busy with my finals so I would send you it later make sure you check it

- 2 years, 9 months ago

alright, good luck dude

- 2 years, 9 months ago

@P C Thanks, but please check it out later, but it may contain flaws and feel free to spot the mistakes

- 2 years, 9 months ago

How did you proved the first part

- 2 years, 9 months ago

First can be proved easily using AM-GM inequality.

- 2 years, 9 months ago

Yes just saw that

- 2 years, 9 months ago