# 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
3 years, 7 months ago

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How did you proved the first part

- 3 years, 7 months ago

First can be proved easily using AM-GM inequality.

- 3 years, 7 months ago

Yes just saw that

- 3 years, 7 months ago

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

- 3 years, 7 months ago

can you show me how?

- 3 years, 7 months ago

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

- 3 years, 7 months ago

yes I am, can you send it to me?

- 3 years, 7 months ago

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

- 3 years, 7 months ago

alright, good luck dude

- 3 years, 7 months ago

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

- 3 years, 7 months ago