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?

#Algebra

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