How can I prove the following inequality
$$\frac{2a}{1+b^2}+\frac{2b}{1+c^2}+\frac{2c}{1+a^2}\geq 3, \forall\ a,b,c>0, a+b+c=3.$$
I tried Cauchy inequality, AM-GM, but I don't get anything good...
2026-03-30 06:43:54.1774853034
Cyclic Inequality in 3 variables
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By AM-GM $\sum\limits_{cyc}\frac{2a}{1+b^2}=\sum\limits_{cyc}\left(\frac{2a}{1+b^2}-2a\right)+6=\sum\limits_{cyc}\frac{-2ab^2}{1+b^2}+6\geq\sum\limits_{cyc}\frac{-2ab^2}{2b}+6=$ $=6-(ab+ac+bc)\geq6-3=3$