<s>[INST] Prove the inequality $\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$ For every $a,b,c>0$ such that $abc=1$ prove the inequality 
$$\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$$
My work so far:
$abc=1 \Rightarrow \frac 13 (a+b+c)\ge 1$
Then $$c^2+a+b \le c^2+(a+b)\frac 13(a+b+c)$$
Or more
$$ (\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a})((c^2+a+b)+(a^2+b+c)+(b^2+c+a)) \ge $$
$$\ge(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2$$
 [/INST] As we know:
$$2ab \le a^{2}+b^{2} $$, and
$$3a^{\frac{1}{3}} = 3(a^{2}bc)^{\frac{1}{3}} \le a^{2} + b + c $$ (because of $abc =1$)
Now we have :
$$2(\frac{ab}{3c^{1/3}} + \frac{ac}{3b^{1/3}} +\frac{cb}{3a^{1/3}})$$
 Now using Cauchy we could prove our inequality.
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