Prove the inequality $\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$

Question

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 $$ (\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a})((c^2+a+b)+(a^2+b+c)+(b^2+c+a)) \ge $$ $$\ge(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2$$

Answer 1

Let $a=x^3$, $b=y^3$ and $c=z^3$. Hence, by Muirhead $\sum\limits_{cyc}\frac{a^2+b^2}{c^2+a+b}=\sum\limits_{cyc}\frac{x^6+y^6}{z^6+xyz(x^3+y^3)}\geq\sum\limits_{cyc}\frac{x^6+y^6}{z^6+z(x^5+y^5)}=\frac{\sum\limits_{cyc}(x^7y+x^7z)}{xyz(x^5+y^5+z^5)}\geq2$

Answer 2

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.